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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Show that the points `A(1,-2,-8),B(5,0,-2)a n dC(1,3,7)`are collinear, and find the ratio in which `B`divides `A Cdot` |
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Answer» The position vectors of A, B and C are `(hati - 2hatj - 8hatk), (5hati-2hatk)` and `(11hati + 3hatj + 7hatk)` respectively. `:. vec(AB) =` (position vector of B) - (position vector of A) `= (5hati - 2hatk( - (hati - 2hatj +8hatk) = (4hati + 2hatj + 6hatk)`. `vec(BC) =` (position of C) - (position vector of B) `= (11hati + 3hatj + 7hatk) - (5hati - 2hatk) = (6hati+3hatj +9hatk)`, and `vec(AC) =` (position vector of C) - (position vector of A) `(11hati + 3hatj + 7hatk) - (-hati - 2hatj - 8hatk) = (10 hati + 5hatj + 15 hatk)`. Now, `vec(AB) = (4hati + 2hatj + 6hatk) = 2(2hati + hatj + 3hatk)`. `= 2/10 (6 hati + 3hatj + 9hatk) = 2/3 vec(BC)`. `:. (|vec(AB)|)/(|vec(BC)|) = 2/3` Hence, B divides AC in the ratio `2:3`. |
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| 2. |
Find the position vector of the mid point of the vector joining thepoints `P(2,3,4)a n d Q(4,1,-2)dot` |
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Answer» The position vectorsofthe given points P and Q are `veca = (2veci + 3vecj + 4veck)` and `vecb= (4hati + hatj - 2hatk)` respectively. |
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| 3. |
Classify the following measures as scalars and vectors : (i) `40` seconds , (ii) `100 m^(2)`, (iii) `30 gm//cm^(3)` (iv) `60 km//hr` , (v) `56 m//s` towards south |
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Answer» (i) `40` seconds represents time, which is scalar. `100 m^(2)` represents an area, which is scalar. (iii) `30 gm//cm^(3)` represents density, which is scalar. (iv) `60 km//hr` represents speed, which is scalar, (v) `56m//s` towards south represents velocity, which is a vector. |
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| 4. |
Write the different vectors having same magnitude. |
| Answer» Clearly, `3vec(AB)` and `5 veca(AB)` are two different values having the same direction. | |
| 5. |
Write down the magnitude of each of the following vectors : (i) `veca = hati + 2hatj + 5hatk` , (ii) `vecb = 5hati - 4hatj - 3hatk` (iii) `vecc = (1/(sqrt(3))hati - (1)/(sqrt(3)) hatj + 1/(sqrt(3))hatk)` , (iv) `vecd = (sqrt(2)hati + sqrt(3)hatj - sqrt(5) hatk)` |
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Answer» Correct Answer - (i) `sqrt(30)` ,(ii) `5sqrt(2)`, (iii) 1, (iv) `sqrt(10)` |
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| 6. |
Find a vector of magnitude 21 units in the direction of the vector `(2hati - 3hatj + 6hatk)`. |
| Answer» Correct Answer - `(6hati - 9hatj + 18 hatk)` | |
| 7. |
Find a vector of magnitude 8 units in the direction of the vector `(5hati - hatj + 2hatk)`. |
| Answer» Correct Answer - `8/(sqrt(30)) (5hati - hatj + 2hatk)` | |
| 8. |
Find a vector of magnitude 9 units in the direction of the vector `(-2hati + hatj + 2hatk)`. |
| Answer» Correct Answer - `(-6hati + 3hatj + 6hatk)` | |
| 9. |
Find the direction ratios and the direction cosines of the vector joining the points `A(2,1,-2)` and `B(3,5,-4)`. |
| Answer» Correct Answer - `(1,4,-2), (1/(sqrt(21)), 4/(sqrt(21)), (-2)/(sqrt(21)))` | |
| 10. |
Find the direction ratios and direction cosines of the vector `veca = (5hati - 3hatj + 4hatk)`. |
| Answer» Correct Answer - `(5,-3,4), (1/(sqrt(2)) , (-2)/(5sqrt(2)), (4)/(5sqrt(2)))` | |
| 11. |
Find a unit vector in the direction of vector `vecb = hati + 2hatj + 3hatk`. |
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Answer» `veca = hati + 2hatj + 3hatk rArr |veca| = sqrt(1^(2) + 2^(2) + 3^(2)) = sqrt(14)`. Unit vector in the direction of `veca` is given by `hati = (veca)/(|veca|) = ((hati +2hatj + 3hatk))/(sqrt(14)) = ((1)/(sqrt(14)) hati + 2/(sqrt(14)) hatj + 3/(sqrt(14)))` |
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| 12. |
Findthe unit vector in the direction of vector ` -> P Q`,where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively. |
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Answer» We have, p.v. of `A = (hati + 2hatj + 3hatk)` and p.v.`B = (4hati + 5hatj + 6hatk)`. `:. vec(AB) = (p.v. of B) - (p.v. of A)` `= (4hati + 5hatj + 6hatk) - (hati + 2hatj + 3hatk) = (3hati + 3hatj + 3hatk)`. And `|vec(AB)| = sqrt(3^(2) + 3^(2) + 3^(2)) = sqrt(27)`. `:.` unit vector in the direction `vec(AB)` . `= (vec(AB))/(|vec(AB)|) = (3hati + 3hatj + 3hatk)/(sqrt(27)) = (3(hati + hatj + hatk))/(3sqrt(3)) = ((hati + hatj + hatk))/(sqrt(3))` `= (1/(sqrt(3))hati + 1/(sqrt(3))hatj + 1/(sqrt(3))hatk)`. |
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| 13. |
Let `veca = a_(1)hati+3hatj+a_(3)hatk` and `vecb = 2hati + b_(2)hatj + hatk`. If `veca = vecb`, find the values of `a_(1),b_(2)` and `a_(3)`. |
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Answer» `veca = vecb hArr a_(1)hati + 3hatj +a_(3)hatk = 2hati + b_(2)hatj +hatk` `hArr a_(1) = 2, b_(2) = 3, a_(3) = 1`. |
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| 14. |
Write a vector of magnitude 15 units in the direction of the vector `(hati -2hatk + 2hatk)`. |
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Answer» Let `veca = (hati -2hatj + 2hatk)`. Then a unit vector in the direction of `veca` is given by `hata = (veca)/(|veca|) = ((hati - 2hatj + 2hatk))/(sqrt(1^(2) + (-2)^(2) + 2^(2))) = 1/3(hati - 2hatj + 2hatk)` . `:.` a vector of magnitude 15 in the direction of `veca`. `= 15 xx 1/3 (hati - 2hatj + hatk) = (5hati - 10 hatj + 10 hatk)`. |
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| 15. |
For what value of a, the vector s `(2hati - 3hatj + 4hatk)` and `(ahati + 6hatj - 8hatk)` collinear ? |
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Answer» The given vectors are collinear only when `(ahati + 6hatj - 8hatk) = lambda(2hati - 3hatj + 4hatk)` for some nonzero scalar `lambda` . Now, `ahati + 6hatj - 8hatk = 2lambda hati - 3lambda hatj + 4lambdahatk` `hArr 2lambda = a , - 3lambda = 6` and `4lambda = - 8`. `hArr = a/2` and `lambda = - 2 hArr a/2= - 2 hArr a = - 4` Hence the given vectors are collinear when `a = - 4`. |
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| 16. |
Find a unit vector parallel to the sum of the vector `(hati + hatj + hatk)` and `(2hati - 3hatj + 5hatk)`. |
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Answer» Let `veca = (hati hatj + hatk)` and `vecb = (2hati - 3hatk + 5hatk)`. Then, `(veca+vecb) = (hati + hatj + hatk) + (2hati - 3hatj + 5hatk) = (3hati - 2hatj + 6hatk)` Required unit vectors parallel to `(veca + vecb)` are `+-((3hati - 2hatj + 6hatk))/(sqrt(3^(2) + (-2)^(2) + 6^(2))) = +- 1/7(3hati - 2hatj + 6hatk)`. |
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| 17. |
If `veca = (2hati - 4hatj + 5hatk)` then find the value of `lambda` so that `lambdaveca` may be a unit vector. |
| Answer» Correct Answer - `+- 1/(3sqrt(5))` | |
| 18. |
Show that the three points `-2hati+3hatj+5hatk, hati+2hatj+3hatk, 7hati-hatk` are collinear |
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Answer» Clearly,we have `vec(AB)` = (position vector of B) - (position vector of A) `= (hati + 2hatj + 3hatk) - (-2hati + 3hatj + 5hatk) = (3hati - hatj - 2hatk)` `vec(BC) =` (position vector of C) - (position vector of B) `= (7hati - 3hatk) - (hati + 2hatj +3hatk) =(6hati - 2hatj - 4hatk)` `:. vec(AB) = 2vec(BC)` , which shows that `vec(AB) ` and `vec(BC)` are parallel vectors, Hence the points A,B and C collinear. |
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| 19. |
Show that the points `A,B` and C having position vectors `(hati + 2hatj + 7hatk),(2hati + 6hatj + 3hatk)`, and `(3hati + 10 hatj - 3hatk)` respectively, are collinear. |
| Answer» Correct Answer - N//A | |
| 20. |
If `veca = (hati - 2hatj), vecb = (2hati - 3hatj)` and `vecc = (2hati + 3hatk)`, find `(veca + vecb + vecc)`. |
| Answer» Correct Answer - `(5hati - 5hatj + 3hatk)` | |
| 21. |
Write the direction cosines of the vectors `(-2hati + hatj - 5hatk)`. |
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Answer» The given vector is `veca = (-2hati + hatj -5hatk)`. Direction ratios of `veca` are `-2,1,-5`. `|veca| = sqrt((-2)^(2) + 1^(2) +(-5)^(2)) = sqrt(30)`. Hence, the direction cosines of `veca` are `(-2)/(sqrt(30)), (1)/(sqrt(30)), (-5)/(sqrt(30))`. |
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| 22. |
If `vec(AB) = (2hati + hatj - 3hatk)` and `A(1,2,-1)` is the given point, find the coordinates of B. |
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Answer» Correct Answer - `(3,3,-4)` Let O be the origin. Then, `vec(OA) = (hati + 2hatj - hatk)`. `:. vec(AB) = (vec(OB) - vec(OA)) rArr vec(OB) = (vec(AB) + vec(OA))`. |
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| 23. |
What is the cosine of the angle which the vector `sqrt(2) hat i+ hat j+ hat k`makes with y-axis? |
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Answer» The given vector is `veca = (sqrt(2) hati + hatj + hatk)` . Direction ratios of `veca` are `sqrt(2),1,1`. `|veca|= sqrt((-2)^(2) + 1^(2) + 1^(2)) = sqrt(4) = 2` `:.` direction cosines of `veca` are `(sqrt(2))/(2), 1/2,1/2` Let `veca` make angle `beta` with the y-axis. Then, clearly, `cos beta = 1/2`. |
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| 24. |
If `A(-2,1,2)` and `B(2,-1,6)` are two given point, find a unit vector in the direction of `vec(AB)`. |
| Answer» Correct Answer - `(2/3 hati - 1/3 hatj + 2/3 hatk)` | |
| 25. |
Find the values of x for which `x(hati+hatj+hatk)` is a unit vector |
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Answer» `p(hati + hatj + hatk)` is a unit vector. `hArr |p(hati + hatj + hatk)|^(2) = 1 hArr(p^(2) +p^(2) + p^(2)) = 1` `hArr 3p^(2) = 1 hArr p^(2)= 1/3 hArr p = +- 1/(sqrt(3))`. |
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| 26. |
Write a unit vector in the direction of `vec(PQ)` , where P and Q are the points `(1,3,0)` and `(4,5,6)` respectively. |
| Answer» Correct Answer - `1/7(3hati + 2hatj + 6hatk)` | |
| 27. |
If `A(1,2,-3)` and `B(-1,-2,1)` are the two given points in space then find (i) the direction ratios of `vec(AB)` and (ii) the direction cosines of`vec(AB)`. Express `vec(AB)` in terms of `hati , hatj` and `hatk`. |
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Answer» (i) The direction ratios of `vec(AB)` are `(-1,-1),(-2,-2),(1+3)` , i.e, `-2,-4,4`. (ii) `|vec(AB)| = sqrt((-2)^(2) + (-4)^(2) + 4^(2)) = sqrt(36) = 6`. `:.` the direction cosines of `vec(AB)` are `(-2)/(6), (-4)/(4),4/6` i.e., `(-1)/(3),(-2)/(3), (2)/(3)`. Clearly, `vec(AB) = - 1/3 hati, - 2/3 hatj + 2/3 hatk`. |
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| 28. |
Writetwo different vectors having same magnitude. |
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Answer» Consider the vectors `veca = 2hati + 3hatj - 4hatk` and `vecb = 4hati + 2hatj + 3hatk`. Clearly, `veca ne vecb` . But `|veca| = sqrt(2^(2) + 3^(2) +(-4)^(2)) = sqrt(29)` and `|vecb| = sqrt(4^(2) + 2^(2) + 3^(2)) = sqrt(29)`. Thus, `|veca| = |vecb|` and `veca ne vecb`. |
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| 29. |
Find the scalar and scalar components of the vector with initial points `A(-3,-1,2)` and terminal point `B(-5,4,3)`. |
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Answer» We have, `vec(AB) =` (position of vector) - (position vectors of A) ` = (-5hati + 4hatj + 3hatk) - (3hati - hatj + 2hatk) = (-8hati+ 5hatj + hatk)`. The scalar components of `vec(AB)` are `-8,5,1`. The vector component of `vec(AB)` are `-8hati 5hatj , hatk`. |
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| 30. |
Using vector method, show that the points `A(1,-1,0), B(4,-3,1)` and `C(2,-4,5)` are the vertices of a right -angled triangle. |
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Answer» The position vectors of A,B,C are `(hati - hatj),(4hati - 3hatj + hatk)` and `(2hati - 4hatj + 5hatk)` respectively. Show that `vec(AB) + vec(BC) + vec(CA) = vec0` and `|vec(AB)|^(2) + |vec(BC)|^(2) = |vec(CA)|^(2)`. `:. DeltaABC` is right angled at B. |
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| 31. |
Find the position vector of the mid-point of the vector joining the points `A(3hati + 2hatj + 6hatk)` and `B(hati + 4hatj - 2hatk)` |
| Answer» Correct Answer - `P (2hati + 3hatj + 2hatk)` | |
| 32. |
Find the position vector of a point `R`whichdivides the line joining two points `P`and `Q`whoseposition vectors are `(2 vec a+ vec b)`and (` vec a- 3 vec b)`respectively, externally in the ratio1:2.Also, show that `P`is themid-point of the line segment `R Qdot` |
| Answer» Correct Answer - `(3veca+ 5vecb)` | |
| 33. |
Find the position vector of the point which divides the join of the points `(2veca - 3vecb)` and `(3veca - 2vecb)` (i) internally and (ii) externally in the ratio `2 : 3` . |
| Answer» Correct Answer - (i) `12/5veca - 13/5 vecb ` , (ii) `- 5vecb` | |
| 34. |
Find the position vector of a point R which divides the line segment joiningthe points `A(2,-3,4)` and `B(3,1,-2)` externally in the ratio `3:2`. |
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Answer» The position of `A` is `(2hati -3hatj + 4hatk)`. The position vector of `B` is `(3hati + hatj - 2hatk)`. Let R divide AB extenally in the ratio `3:2` . Then, position vector of R. `= ((3vecb - 2veca)/(3-2))= (3(3hati + hatj - 2hatk) - 2(2hati - 3hatj + 4hatk))/(1)` `= (5hati + 9hatj - 14 hatk)`. Hence, the position vector of R is `(5hati + 9hatj - 14hatk)`. |
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| 35. |
Prove that the points `2hati-hatj+hatk, hati-3hatj-5hatk and 3hati-4hatj-4hatk ` are the vertices of a right angled triangle. Also find the remaining angles of the triangle. |
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Answer» We have, `vec(AB) =` (position of vector) - (position vectors of A) ` = (vecb - veca) = (2hati - hatj + hatk) - (3hati - 4hatj - 4hatk)` `= (-hati + 3hatj + 5hatk)`. `vec(BC) =` (position vector of C) - (position vector of B) `=(vecc - vecb) = (hati - 3hatj - 5hatk) - (2hati - hatk + hatk)`. `= (-hati - 2hatj - 6hatk)`. `vec(CA) =` (position of vector of A) - (position of vector B). ` = (veca - vecc) = (3hati - 4hatj - 4hatk) - (hati - 3hatj - 5hatk)`. `= (2hati - hatj + hatk)`. `:. |vec(AB)|^(2) - {(-1)^(2) + 3^(2) + 5^(2)} = (1+9+25) = 35`, `|vec(BC)|^(2) = {(-1)^(2) + (-2)^(2) + (-6)^(2) } = (1+4+36) = 41`. `|vec(CA)| = {2^(2) + (-1)^(2) + 1^(2)} = (4+1+1) = 6`. `:. |vec(AB)|^(2)= |vec(CA)|^(2) = |vec(BC)|^(2) rArr AB^(2) + CA^(2) = BC^(2)`. Hence, `DeltaABC` is right-angled at A. |
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| 36. |
`P`and `Q`are twopoints with position vectors `3 vec a-2 vec b`and ` vec a+ vec b`respectively. Write theposition vector of a point R which divides the line segment PQ in the ratio2:1 externally. |
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Answer» The position vectors of the given points are `P(3veca - 2vecb)` and `Q(veca+vecb)` We have to divide PQ in the ratio `2:1` externally at the points R. The position vector of R is `(2(veca+vecb)-1.(3veca-2vecb))/((2-1)) = (-veca+4vecb)`. Hence, the position vector of `R` is `(-veca + 4vecb)`. |
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| 37. |
Find the position vector of a point R which divides the line joining the point `P(hati + 2hatj - hatk)` and `Q(-hati + hatj + hatk)` in the ratio `2 : 1`, (i) internally and (ii) externally. |
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Answer» Here`veca = (hati + 2hatj - hatk)` and `vecb= (-hati + hatj + hatk)`. Also `m = 2, n = 1`. (i) When R divides PQ internally in the ratio `2 : 1`, then position vector of `R = ((m vecb + n veca))/((m+n))` `= (2(-hati + hatj + hatk) + 1.(hati + 2hatj - hatk))/((2+1))` (ii) When R divides PQ externally in the ratio`2 : 1`, then position vector of `R = ((mvecb - nveca))/((m-n))` `= (2(-hati + hatj + hatk) + 1.(hati + 2hatj - hatk))/((2-1))` `= (-3hati + 3hatk)`. |
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| 38. |
If `veca = (3hati + hatj - 5hatk)` and `vecb = (hati + 2hatj - hatk)` then find a unit vector in the direction of `(veca-vecb)`. |
| Answer» Correct Answer - `1/(sqrt(21)) (2hati - hatj - 4hatk)` | |
| 39. |
If `veca = (-hati + hatj - hatk)` and `vecb = (2hati- 2hatj + 2hatk)` then find the unit vector in the direction of `(veca + vecb)`. |
| Answer» Correct Answer - `1/(sqrt(2)) (hati + hatk)` | |
| 40. |
If the position vector of the vertices A,B and C of a `DeltaABC` be `(hati + 2hatj + 3hatk), (2hati + 3hatj + hatk)` and `(3hati + hatj +2hatk)` respectively, prove that `DeltaABC` is equilateral. |
| Answer» Show that `|vec(AB)|= |vec(BC)| = |vec(CA)|` | |
| 41. |
If `veca = (hati + hatj + hatk), vecb = (4hati - 2hati + 3hatk)` and `vecc = (hati - 2hatj + hatk)`, find a vector of magnitude 6 units which is parallel to the vector `(2veca - vecb + 3vecc)`. |
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Answer» We have, `(2veca - vecb + 3vecc) = 2(hati + hatj + hatk) - (4hati - 2hatj + 3hatk) + 3(hati-2hatj +hatk)` `= (2-4+3) hati + (2+2-6)hatj + (2-3+3) hatk` `= (2hati - 2hatj + 2hatk)`. Unit vectors parallel to `(2veca - vecb + 3vecc)` are `+- ((hati - 2hatj + 2hatk))/(sqrt(1^(2) + (-2)^(2) + 2^(2))) = +- 1/3(hati - 2hatj + 2hatk)` . Required vectors of magnitude of 6 units are `+-{6xx 1/3(hati - 2 hatj+ 2hatk)} = +- 2(hati - 2hatj + 2hatk)`. |
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| 42. |
Find the position vector of a point R which divides the line joining `A(-2,1,3)` and `B(3,5,-2)` in the ratio `2 : 1` (i) internally (ii) externally. |
| Answer» Correct Answer - (i) `{4/3hati + 11/3hatj - 1/3 hatk}`, (ii) `(8hati + 9hatj - 7 hatk)` | |