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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
When two vectors of magnitudes P and Q are inclined at an angle `theta`, the magnitudes of their resultant is 2P. When the inclination is changed to `180^(@)-theta`, the magnitudes of the resultant is halved. Find the ratio of `P` and `Q`. |
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Answer» Correct Answer - `sqrt(2)/sqrt(3)` `R_(1)^(2)=(2P)^(2)=P^(2)+Q^(2)+2PQ cos theta` `(P)^(2)=P^(2)+Q^(2)-2PQ cos theta` Adding them `5P^(2)=2P^(2)+2Q^(2)` or `3P^(2)=2Q^(2)` or `P/Q=sqrt(2)/sqrt(3)` |
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| 2. |
If the resultant of two forces of magnitudes `p` and `2p` is perpendicular to `p`, then the angle between the forces isA. `(2pi)/3`B. `(3pi)/4`C. `(4pi)/5`D. `(5pi)/6` |
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Answer» Correct Answer - A `tan theta= (2p sin theta)/(p+2p cos theta), here alpha = 90^(@)` `implies oo=(2p sin theta)/(p+2p cos theta)implies p+2p cos theta=0 `implies `cos theta-1/2implies theta= (2pi)/3` |
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| 3. |
In a methane `(CH_(4)` molecule each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the centre. In coordinates where one of the `C-H` bond in the `hat(i)-hat(j)-hat(k)`, an adjacent `C-H` bond in the `hat(i)-hat(j)-hat(k)` direction. Then angle between these two bonds.A. `cos^(-1)(2/3)`B. `cos^(-1)(2/3)`C. `cos^(-1)(-1/3)`D. `cos^(-1)(1/3)` |
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Answer» Correct Answer - C `cos theta=((hat(i)+hat(j)+hat(k)).(hat(i)-hat(j)-hat(k)))/(sqrt(3))= (1-1-1)/3= -1/3` |
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| 4. |
If `vecA,vecB,vecC` are mutually perpendicular show that `vecCxx(vecAxxvecB)=0`. Is the converse true? |
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Answer» `AxxB` is perpendicular to both A and `B` so this is parallel or antiparallel to `C`. Now, cross, product of two parallel or antiparallel vectors is zero. Hence, `Cxx(AxxB) = 0`. |
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| 5. |
For the any two vecrtors `vecA` and `vecB`, if `vecA.vecB=|vecAxxvecB|`, the magnitude of `vecC=vecA+vecB` is equal toA. `sqrt(A^(2)+B^(2)`B. `A+B`C. `sqrt(A^(2)+B^(2)+(AB)/(sqrt(2)))`D. `sqrt(A^(2)+B^(2)+sqrt(2)xxAB)` |
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Answer» Correct Answer - D |
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| 6. |
A magnitude of vector `vecA,vecB` and `vecC` are respectively `12, 5` and `13` units and `vecA+vecB=vecC` then the angle between `vecA` and `vecB` isA. `0`B. `pi`C. `pi//2`D. `pi//4` |
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Answer» Correct Answer - C |
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| 7. |
A magnitude of vector `vecA,vecB` and `vecC` are respectively `12, 5` and `13` units and `vecA+vecB=vecC` then the angle between `vecA` and `vecB` is |
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Answer» Correct Answer - C `|vecA + vecB|=|vecC|` `sqrt(A^(2) + B^(2) + 2AB cos theta)= sqrt(13^(2))` `12^(2) + 5^(2) + 2xx 12 xx 5 cos theta = 13^(2)` `2 xx 12 xx 5 cos theta = 169 - 169 =0` `" "cos theta= 0` `" "theta =(pi)/(2)` |
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| 8. |
If `vecA = vecB + vecC` and the magnitudes of `vecA, vecB and vecC` are 5,4 and 3 units respecetively, the angle between `vecA` and `vecC` is :A. `cos ^(-1)(3//5)`B. `cos ^(-1)(4//5)`C. `pi//2`D. `sin^(-1)(3//4)` |
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Answer» Correct Answer - A `vecC + vecB = vecA` `rArr " "vecB = vecA - vecC` `" "|vecB|=|vecA - vecC|` `" "4=sqrt(5^(2) + 3^(2) - 30 cos theta)` `" "16-34 =-30cos theta` `" "(-18)/(-30) =cos theta" "rArr" "cos theta= (3)/(5)` `" " theta = cos ^(-1).(3)/(5)` |
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| 9. |
Two vectors `vec A and vecB` have equal magnitudes.If magnitude of `(vecA+vecB)` is equal to n times ofthe magnitude of `(vecA-vecB)` then the anglebetween `vecA and vecB` is :-A. `cos^(-1) ((n - 1)/(n + 1))`B. `cos^(-1) ((n^(2) - 1)/(n^(2) + 1))`C. `sin^(-1) ((n - 1)/(n + 1))`D. `sin^(-1) ((n^(2) - 1)/(n^(2) + 1))` |
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Answer» Correct Answer - B Let `theta` be the angle between `vec(A)` and `vec(B)` `|vec(A) + vec(B)| = n|vec(A) - vec(B)| rArr sqrt(A^(2) + B^(2) + 2 AB cos theta)` `= n sqrt(A^(2) + B^(2) + 2 AB cos (180 - theta))` `|vec(A)| = |vec(B)| = A = B = x` `2x^(2) ( 1 + cos theta) = n^(2) . 2x^(2) ( 1 - cos theta)` `1 + cos theta = n^(2) . n^(2) cos theta` `(1 + n^(2)) cos theta = n^(2) - 1` `cos theta = (n^(2) - 1)/(n^(2) + 1)` `theta = cos^(-1) ((n^(2) - 1)/(n^(2) + 1))` |
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| 10. |
Consider three vectors `vecA, vecB and vecC` having magnitudes 4, 5, and 3. These vectors are of similar nature, e.g. these could be there displacement. Apply your answer understanding of vectors algebra to match Column-I with Column-II. |
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Answer» Correct Answer - `a to r; b to p; c tos; d to q` `|vecA|= 4 , |vecB|=5, |vecC|=3` |
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| 11. |
A particle is initially at point `A(2,4,6)m` moves finally to the point` B(3,2,-3)m`. Write the initial position vector,final position,and displacement vector of the particle. |
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Answer» Initial position vector : `vec(r )_(1)=2hat(i)+4hat(j)+6hat(k)` Final position vector : `vec(r )_(2)=3hat(i)+2hat(j)-3hat(k)` Displacement : `vec(d)=vec(r )_(2)-vec(r )_(1)=(3-2)hat(i)+(2-4)hat(j)+(-3-6)hat(k)` `=hat(i)-2hat(j)-9hat(k)` |
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| 12. |
If the resultant of `n` forces of different magnitudes acting at a point is zero, then the minimum value of `n` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C If Vectors are of equal magnitudes then two vectors can give zero resultant, if they works in opposite direction. But if the vector are of different magnitudes tehn minimum three vectors are required to give zero resultant. |
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| 13. |
If a line makes angles 90º, 135º, 45º with the X, Y and Z axes respectively, then find its direction cosines. |
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Answer» Let l, m, n be the direction cosines of the line. Then l = cos α, m = cos β, n = cos γ Here, α = 90°, β = 135° and γ = 45° ∴ l = cos 90° = 0 m = cos 135° = cos (180° – 45°) = -cos 45° \(= -\frac{1}{\sqrt{2}}\) and n = cos 45° \(= \frac{1}{\sqrt{2}}\) ∴ the direction cosines of the line are 0, \(-\frac{1}{\sqrt{2}},\) \(\frac{1}{\sqrt{2}}\) |
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| 14. |
The distance of the point (3, 4, 5) from Y- axis is(A) 3 (B) 5(C) √34(D) √41 |
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Answer» Correct answer is (C) √34 |
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| 15. |
If |a| = 3 and –1 ≤ k ≤ 2, then |ka| lies in the interval(A) [0, 6](B) [-3, 6] (C) [3, 6] (D) [1, 2] |
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Answer» Correct answer is (A) [0, 6] |
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| 16. |
Let vectors p and q be the position vectors of P and Q respectively, with respect to O and |p| = p, |q| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then.(A) 9p2 = 4q2(B) 4p2 = 9q2(C) 9p = 4q (D) 4p = 9q |
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Answer» Correct answer is (A) 9p2 = 4q2 |
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| 17. |
Let α, β, γ be distinct real numbers. The points with position vectors αi + βj + γk, βi + γj + αk, γi + αj + βk(A) are collinear (B) form an equilateral triangle (C) form a scalene triangle (D) form a right angled triangle |
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Answer» Correct answer is (B) form an equilateral triangle |
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| 18. |
If α, β, γ are direction angles of a line and α = 60º, β = 45º, the γ = (A) 30º or 90º (B) 45º or 60º (C) 90º or 30º (D) 60º or 120º |
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Answer» Correct answer is (D) 60º or 120º |
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| 19. |
If cos α, cos β, cos γ are the direction cosines of a line then the value of sin2α + sin2β + sin2γ is(A) 1 (B) 2 (C) 3 (D) 4 |
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Answer» Correct answer is (B) 2 |
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| 20. |
The volume of tetrahedron whose vertices are (1, -6, 10), (-1, -3, 7), (5, -1, λ) and (7, -4, 7) is 11 cu. units then the value of λ is(A) 7(B) π/3(C) 1(D) 5 |
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Answer» Correct answer is (A) 7 |
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| 21. |
The volume (in cubic unit ) of the tetrahedron with edges `hati+hatj+hatk, hati-hatj+hatk and hati+2hatj-hatk`, isA. 4B. `(2)/(3)`C. `(1)/(6)`D. `(1)/(3)` |
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Answer» Correct Answer - B Volume of tetrahedron `=1/6[(AB, AC, AD)]` `=1/6 |(1,1,1),(1,-1,1),(1,2,-1)|` `=1/6 [-1+2+3]=2/3` cu unit |
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| 22. |
Obtain the magnitude and direction cosines of vector `(A-B),` if `A=2hati+3hatj+hatk, B=2hati+2hatj+3hatk`A. `0,(1)/(sqrt(5)),(-2)/(sqrt(5))`B. `0,(2)/(sqrt(5)),(1)/(sqrt(5))`C. `0,0,(1)/(sqrt(5))`D. None of these |
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Answer» Correct Answer - A (a) WE have `A-B =hatj-2hatk=C` `C=sqrt(1+4)=sqrt(5)` `cos alpha=(o)/(sqrt(5))=0,cosbeta=(1)/(sqrt(5))and cos gamma =(-2)/(sqrt(5))` |
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| 23. |
Let `D ,Ea n dF`be the middle points of the sides `B C ,C Aa n dA B ,`respectively of a triangle `A B Cdot`Then prove that ` vec A D+ vec B E+ vec C F= vec0`. |
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Answer» Let the position vectors of A, B and C be `veca, vecb and vecc` respectively. Then the positive vectors of D, E and F are `(vecb + vecc)//2, (vecc + veca)//2a and (veca + vecb)//2`, respectively. Therefore, `" "vec(AD) + vec(BE) + vec(CF) = ((vecb + vecc)/(2) - veca) + ((vecc + veca )/(2) - vecb) + ((veca + vecb)/(2) - vecc) = vec0` |
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| 24. |
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6). |
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Answer» Let A = (3, -2, -5), B = (3, -2, 6) The direction ratios of the line AB are 3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11. The parametric equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are x = x1 + aλ, y = y1 + bλ, z = z1 + cλ ∴ the parametric equations of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ i.e. x = 3, y = -2, z = 11λ – 5 ∴ the cartesian equations of the line are x = 3, y = -2, z = 11λ – 5, λ is a scalar. |
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| 25. |
Let `A B C D`be a p[arallelogram whose diagonalsintersect at `P`and let `O`be the origin.Then prove that ` vec O A+ vec O B+ vec O C+ vec O D=4 vec O Pdot` |
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Answer» Since the diagonals of a parallelogram bisect each other, P is the middle point of AC and BD both. Therefore, `vec(OA) + vec(OC) = 2 vec(OP) and vec(OB) + vec(OD) = 2 vec(OP)` |
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| 26. |
The direction cosines of the normal to the plane 2x – y + 2z = 3 are(A) \(\frac{2}{3},\frac{-1}{3},\frac{2}{3}\)(B) \(\frac{-2}{3},\frac{1}{3},\frac{-2}{3}\)(C) \(\frac{2}{3},\frac{1}{3},\frac{2}{3}\)(D) \(\frac{2}{3},\frac{-1}{3},\frac{-2}{3}\) |
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Answer» Correct answer is (A) \(\frac{2}{3},\frac{-1}{3},\frac{2}{3}\) |
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| 27. |
Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is (A) 0º (B) 30º (C) 45º (D) 90º |
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Answer» Correct answer is (A) 0º |
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| 28. |
Show that lines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them. |
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Answer» Given lines are x = y, z = 0 and x + y = 0, z = 0. It is clear that (0, 0, 0) satisfies both the equations. ∴ the lines intersect at O whose position vector is \(\bar{0}\) Since z = 0 for both the lines, both the lines lie in XY- plane. Hence, we have to find equation of XY-plane. Z-axis is perpendicular to XY-plane. ∴ normal to XY plane is \(\hat{k}.\) \(0(\bar{0})\) lies on the plane. By using \(\bar{r} \) . \(\bar{n}\) = \(\bar{a}\) . \(\bar{n}\), the vector equation of the required plane is \(\bar{r} \) . \(\hat{k}\) = \(\bar{0}\) . \(\bar{k}\) i.e. \(\bar{r} \) . \(\hat{k}\) = 0. Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \) . \(\hat{k}\) = 0. |
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| 29. |
The vector equation of line 2x – 1 = 3y + 2 = z - 2 is(a) \(\bar{r} = \left(\frac{1}{2}\bar{i} - \frac{2}{3}\bar{j}+ 2\bar{k}\right)\) + \(\lambda(3\bar{i} + 2\bar{j} + 6\bar{k})\)(b) \(\bar{r} = \bar{i} - \bar{j} + (2\bar{i}+\bar{j} + \bar{k})\)(c) \(\bar{r} = \left(\frac{1}{2}\bar{i} - \bar{j}\right)\) + \(\lambda(\bar{i} - 2\bar{j} + 6\bar{k})\)(d) \(\bar{r} = (\bar{i} + \bar{j})\) + \(\lambda(\bar{i} - 2\bar{j} + 6\bar{k})\) |
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Answer» Correct answer is (a) \(\bar{r} = \left(\frac{1}{2}\bar{i} - \frac{2}{3}\bar{j}+ 2\bar{k}\right)\) + \(\lambda(3\bar{i} + 2\bar{j} + 6\bar{k})\) |
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| 30. |
If the line x/3 = y/4 = z is perpendicular to the line \(\frac{x - 1}{k}=\frac{y + 2}{3}=\frac{z - 3}{k - 1}\) then the value of k is :(x - 1)/k = (y + 2)/3 = (z - 3)/(k - 1)(a) 11/4(b) - 11/4(c) 11/2(d) 4/11 |
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Answer» Correct answer is (b) - 11/4 |
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| 31. |
Assertion: `vec(A)xxvec(B)` is perpendicualr to both `vec(A)-vec(B)` as well as `vec(A)+vec(B)` Reason: `vec(A)+vec(B)` as well as `vec(A)-vec(B)` lie in the plane containing `vec(A)` and `vec(B)`, but `vec(A)xxvec(B)` lies perpendicular to the plane containing `vec(A)` and `vec(B)`.A. If both Assetion and Reason are correct but Reason is the correct explanation of Assertion.B. If both Assetion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Asserion is true but Reason is falseD. If Asserion is false but Reason is true |
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Answer» Correct Answer - A `(a) (AxxB)` is perpendicular to`(A+B) and (A-B)` becasause `(AxxB)` lies perpendicular to the plane containtiing A ns B . |
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| 32. |
If a vector `vec(A)` is parallel to another vector `vec(B)` then the resultant of the vector `vec(A)xxvec(B)` will be equal toA. `A`B. `vecA` and `vecB` act in the same directionC. Zero vectorD. Zero |
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Answer» Correct Answer - C |
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| 33. |
The angle between `vec(A)+vec(B)` and `vec(A)xxvec(B)` is |
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Answer» Correct Answer - C `vec(A)+vec(B)` will be in the plane containing `vec(A)` and `vec(B)`, whereas `vec(A)xxvec(B)` will be perpendicular to that plane. |
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| 34. |
If `vec(A) = hati + 3hatj + 2hatk` and `vec(B) = 3hati + hatj + 2hatk`, then find the vector product `vec(A) xx vec(B)`. |
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Answer» Correct Answer - `4 hati + 4hatj - 8 hatk` |
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| 35. |
A vector `vec(A)` points verically upward and `vec(B)` points towards north. The vector product `vec(A) xx vec(B)` isA. ZeroB. Along westC. Along eastD. Vertically downward |
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Answer» Correct Answer - B `vec(A) xx vec(B) = (A hat(k)) xx (B hat(j)) = - AB hat(i)` |
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| 36. |
The projection of a vector `vec(r )=3hat(i)+hat(j)+2hat(k)` on the `x-y` plane has magnitudeA. 3B. 4C. `sqrt(14)`D. `sqrt(10)` |
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Answer» Correct Answer - D Consider only `x` and `y` components: `sqrt(3^(3)+1^(2))=sqrt(10)` |
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| 37. |
The `x`- component of the resultant of several vectors (i) is equal to the sum of the `x`-components of the vectors (ii) may be smaller than the sum of the magnitudes of the vectors (iii) may be greater than the sum of the magnitudes of the vectors (iv) may be equal to the sum of the magnitudes of the vectorsA. `(i),(ii)`B. `(i) ,(ii),(iv)`C. `(ii) , (iii) , (iv)`D. all |
| Answer» Correct Answer - B | |
| 38. |
A vector `vec(A)` is along the positive `x`- axis . If `B` is another vector such that `vec(A) xx vec(B)` is zero , then `B` could beA. `4 j`B. `- 4 i`C. `-( i + j)`D. `(j + k)` |
| Answer» Correct Answer - B | |
| 39. |
A vector is given by `vec (A) = 3 hat(i) + 4 hat(j) + 5 hat(k)`. Find the magnitude of `vec(A)` , unit vector along `vec(A)` and angles made by `vec(A)` with coordinate axes. |
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Answer» We have ,mangnitude `|A|=A=sqrt(A_(x)^(2)+A_(y)^(2)+A_(z)^(2))` `sqrt((3)^(2)+(4)^(2)+(t)^(2))=5sqrt(2)` `"unit vector,"hatA=(A)/(|A|)=(3hati+4hatj+5hatk)/(5sqrt(2))` Angles made by A with coordinateaxis `cosalpha=(A_(x))/(|A|)=(3)/(5sqrt(2))implies alpha=cos^(-1)((3)/(5sqrt(2)))` `costheta=(A)/(|A|)=(4)/(5sqrt(2))impliesbeta= cos^(-1)((4)/(5sqrt(2)))` `cosgamma=(A_(z))/(|A|)=(5)/(5sqrt(2))` `gamma=cos^(-1)((1)/(sqrt(2)))=(pi)/(4)` |
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| 40. |
The projection of a vector `vec(r )=3hat(i)+hat(j)+2hat(k)` on the `x-y` plane has magnitudeA. 2B. `sqrt(14)`C. `sqrt(10)`D. `sqrt(5)` |
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Answer» Correct Answer - C `vec(R )= 3hat(i)+hat(j)+2hat(k)` `:.` Length in XY plane `= sqrt(R_(x)^(2)+R_(y)^(2))= sqrt(3^(2)+1^(2))= sqrt(10)` |
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| 41. |
If `vec(b)=3hat(i)+4hat(j)` and `vec(a)=hat(i)-hat(j)` the vector having the same magnitude as that of `vec(b)` and parallel to `vec(a)` isA. `5/sqrt(2)(hat(i)-hat(j))`B. `5/sqrt(2)(hat(i)+hat(j))`C. `5(hat(i)-hat(j))`D. `5(hat(i)+hat(j))` |
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Answer» Correct Answer - A Let that vector be `vec(C )`. Then `vec(C )=Chat(C)=bhat(a)=vec(C )=(bhat(a))/a=5/sqrt(2)(hat(i)-hat(j))` |
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| 42. |
If `vec(A) = 3 hat(i) - 4 hat(j)` and `vec(B) = 2 hat(i) + 16 hat(j)` then the magnitude and direction of `vec(A) + vec(B)` will beA. `5 , tan^(-1) (12//5)`B. `10 , tan^(-1) (5//12)`C. `13 , tan^(-1) (12//5)`D. `12 , tan^(-1) (5//12)` |
| Answer» Correct Answer - C | |
| 43. |
A man standing on a road has to hold his umbrella at `30^0` with the vertical to keep the rain away. The throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to a. the road, b. the moving man.A. `10km//hr`B. `20km//hr`C. `30km//hr`D. `40km//hr` |
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Answer» Correct Answer - B |
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| 44. |
if a = (2,1,-1) , b= (1,-1,0) , c = (5,-1,1) then unit vector parallel to a + b -c but opposite directionA. `(1)/(3)(2hati-hatj+2hatk)`B. `(1)/(2)(2hati-hatj+2hatk)`C. `(1)/(3)(2hati-hatj-2hatk)`D. None of these |
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Answer» Correct Answer - A Let a = (2, 1, -1), b = (1, -1, 0) and c = (5, -1, 1) Then, `a+b-c=(2+1-5)hati+(1-1+1)hati+(-1+0-1)hatk` `=-(2hati-hatj+2hatk)` `because` Unit vector of `(a+b-c)=-((2hati-hatj+2hatk))/(3)` `therefore` Required unit vector of `(a+b-c)=((2hati-hatj+2hatk))/(3)` |
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| 45. |
If D, E and F be the middle points of the sides BC,CA and AB of the `DeltaABC`, then `AD+BE+CF` is |
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Answer» Correct Answer - A Let the position vectors A, B, C be a, b and c respectively . Then the position vectors of mid-points D, E and F are ` (b+c)/(2)` ,(c+a)/( 2) and (a+b) /(2)` repectively . Now `AD +BE +CF` `={(b+c)/(2)-a}+{(c+a)/(2)-b}+{(a+b)/(2)-c )}=0` |
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| 46. |
If `vecA = 2hati + hatj -3hatk, vecB = hati- 2hatj + hatk` and `vecC = - hati + hatj - 4hatk`, Calculate (i) `vecA.(vecB xx vecC), (ii)vecC (vecAxx vecB), (iii) vecA xx (vecB xx vecC)` |
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Answer» Correct Answer - (i) 20, (ii) 20,(iii)(`8hati -19hatj+ 3 hatk`) (i) `vecA.(vecBxx vecC) = vecA.[(8-1) hati - (-4 + 1) hatj + (1-2) hatk]` `" "= (2hati + hatj - 3hatk) .(7hati + 3hatj - hatk)=20` (ii) `vecC.(vecA xx vecB) = vecC.[(6-1)hati - (2+3)hatj +(-4-1)hatk]` `" "(-hati + hatj - 4hatk).(5hati -5hatj- 5hatk)= 20` (iii) `vecA xx (vecB xx vecC) = (vecA. vecC) vecB - (vecA. vecB).vecC` `(11) vecB- (3)vecC = (22 hati - 22hatj+ 11hatk)-(3hati - 3hatj- 12hatk)` `=8hati- 19hati -hatk` |
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| 47. |
If `m_(1), m_(2), m_(3)` and `m_(4)` are respectively the magnitudes of the vectors `a_(1)=2hati-hatj+hatk, a_(2)=3hati-4hatj-4hatk, a_(3)=hati+hatj-hatk` and `a_(4)=-hati+3hatj+hatk`, then the correct order of `m_(1),m_(2),m_(3)` and `m_(4)` isA. `m_(3)ltm_(1)ltm_(4)ltm_(2)`B. `m_(3)ltm_(1)ltm_(2)ltm_(4)`C. `m_(3)ltm_(4)ltm_(1)ltm_(2)`D. `m_(3)ltm_(4)ltm_(2)ltm_(1)` |
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Answer» Correct Answer - A Given, `m_(1)=|a_(1)|=sqrt(2^(2)+(-1)^(2)+(1)^(2))=sqrt(6)` `m_(2)=|a_(2)|=sqrt(3^(2)+(-4)^(2)+(-4)^(2))=sqrt(41)` `m_(3)=|a_(3)|=sqrt(1^(2)+1^(2)+(-1)^(2))=sqrt(3)` and `m_(4)=|a_(4)|=sqrt((-1)^(2)+(3)^(2)+(1)^(2))=sqrt(11)` `therefore m_(3)ltm_(1)ltm_(4)ltm_(2)` |
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| 48. |
If `a+b+c=0` and `|a|=5, |b|=3` and `|c|=7`, then angle between a and b isA. `pi/2`B. `pi/3`C. `pi/4`D. `pi/6` |
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Answer» Correct Answer - B Given, `|a|=5, |b|=3, |c|=7 and a+b+c=0` `impliesa+b=-c` On squaring both sides, we get `(a+b)^(2)=(-c)^(2)implies|a+b|^(2)=|c|^(2)` `implies|a|^(2)+|b|^(2)+2a*b=|c|^(2)` `[because theta` be the angle between a and b] `implies(5)^(2)+(3)^(2)+2|a||b|cos theta =(7)^(2)` `implies25+9+2.5.3cos theta =49` `implies30 cos theta =15` `implies cos theta =1/2=cos 60^(@)` `impliestheta =(pi)/(3)` |
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| 49. |
The position vectors of the vertices `A, B and C` of triangle are `hati+hatj, hatj+hatk and hati+hatk`, respectively. Find the unit vectors `hatr` lying in the plane of `ABC` and perpendicular to `IA`, where I is the incentre of the triangle. |
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Answer» Correct Answer - `vecr = pm (1)/(sqrt2)(hati +hatj)` Since `|vec(AB)|= |vec(BC)|= |vec(CA)|`, the incentre is same as the circumcentre, and hence IA is perpendicular of BC. Therefore, `vecr` is parallel to BC. `vecr= lamda (hati-hatj)` Hence, unit vector `vecr=pm (1)/(sqrt2)(hati-hatj)` |
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| 50. |
Let ABC be a triangle the position vectors of whose vertices are respectively `hati+2hatj+4hatk, -2hati+2hatj+hatk and 2hati+4hatj-3hatk`. Then the `/_ABC` is (A) isosceles (B) equilateral (C) righat angled (D) none of theseA. isoscelesB. equilateralC. right angledD. none of these |
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Answer» Correct Answer - C `vec(BC) = vec(OC)-vec(OB) = 4hati+2hatj-4hatk` `" "vec(AB) = -3hati-3hatk, vec(AC) = hati=2hatj- 7hatk` `" "BC^(2) = 36, AB^(2) = 18, AC^(2)= 54 ` Cleary, `AC^(2) = BC^(2)m + AB^(2)` `therefore angleB= 90^(@)` |
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