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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. |
If `alpha,beta ,gamma` are direction angles of a line , then `cos 2alpha + cos 2beta + cos 2 gamma = `A. `-1`B. 0C. 1D. 2 |
| Answer» Correct Answer - a | |
| 2. |
Find the perpendicular distance from the point `(1,-3,4)` to the plane `3x-4y+12z - 1 = 0`. |
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Answer» Perpendicular distance from point `(1,-3,4)` to the plane `3x-4y+12z-1 = 0` is `|(3(1)-4(-3)+12(4)-1)/(sqrt(3^(2)+(-4)^(2)+12^(2)))|` `= |(3+12+48-1)/(13)|=62/13` units. |
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| 3. |
The direction ratios of vector perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4 , 1 areA. `-3, 5, 11 `B. `3, -5, 11`C. `3, 5, 11 `D. `3, 5, -11` |
| Answer» Correct Answer - A | |
| 4. |
If direction ratios of two lines are `2,-6,-3 and 4,3,-1` then direction ratios of a line perpendicular to them areA. `2,3,3`B. `3,-2,6`C. `1,2,3`D. `2,-3,6` |
| Answer» Correct Answer - b | |
| 5. |
Show that the three lines with direction cosines `12/13,-3/13,-4/13,4/13,12/13,3/13,3/13,-4/13,12/13 ` are mutually perpendicular. |
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Answer» (i) For first two lines ` l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)` `= 12/13 xx 4/13 + (-3/13) xx 12/13 + (-4/13) xx 3/13` `= 48/169- 36/169 - 12/169 = 0` Therefore , two lines are perpendicular (ii) For second and third line, `l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)` ` = 4/13 xx 3/12 + 12/13 xx ((-4)/(13)) + 3/13 xx 12/13` `=12/169-48/1639+36/169=0` Therefore, two lines are perpendicular (iii) For third and first line `l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)` `= 3/13 xx 12/13+((-4)/(13)) xx ((-3)/(13)) + (12)/(13) xx ((-4)/(13))` `= 36/169+12/169-48/169=36/169-36/169=0` Therefore, two lines are perpendicular. Hence, the lines are mutually perpendicular. |
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| 6. |
If `bar(a),bar(b),bar (c )` are pair wise mutually perpendicular vectors of equal magnitude , then the angle wisv `bar(a)+bar(b)+bar (c )` isA. `cos ^(-1)(1/(sqrt(3)))`B. `cos^(-1)(1/3)`C. `cos^(-1)(2/(sqrt(3)))`D. `cos^(-1)(sqrt(3))` |
| Answer» Correct Answer - a | |
| 7. |
Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane. |
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Answer» Here, we are given points `A(3,4,1)` and `B(5,1,6)`. So, direction ratio of line `AB = (2,-3,5)`. So, equation of line becomes, `(x-3)/2 = (y-4)/-3 = (z-1)/5 = k` `=>(z-1)/5 = k` As the line is crossing XY-plane, z-coordinate will be `0`.So, `(0-1)/5 = k => k = -1/5` `:. (x-3)/2 = -1/5 => x = 3-2/5 = 13/5` `(y-4)/-3 = -1/5 => y = 4+3/5 = 23/5` So, the required coordinates will be `(13/5,23/5,0)`. |
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| 8. |
If a line has direction ratios `2,1,2`.determine its direction cosines. |
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Answer» The direction cosines of the lines are`(2/3,-1/3,-2/3)`. |
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| 9. |
Direction cosines of the line passing through `A(2,3,-1)and B(-3,-4,2)` areA. `(-5)/(sqrt(83)),(-7)/(sqrt(83)),3/(sqrt(83))`B. `5/(sqrt(83)),7/(sqrt(83)),3/(sqrt(83))`C. `(-7)/(sqrt(83)),3/(sqrt83),(-5)/(sqrt(83))`D. `-5,-7,3` |
| Answer» Correct Answer - a | |
| 10. |
Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line `2x + y = 5`. Then the area of the triangle is |
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Answer» `AD=|(0+0-5)/sqrt(2^2+1^2)|=5/sqrt5=sqrt5` in `/_ABD` `tan 45^o=(AD)/(BD)=sqrt5/(BD)=1` BD=`sqrt5` so, DC=`sqrt5` BC=DC+`sqrt5` BC=`sqrt5+sqrt5` BC=`2sqrt5` area of `/_ABC = 1/2*B*H` =`1/2*BC*AD` =`1/2*2sqrt5*sqrt5` =`5units^2` |
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| 11. |
Find the direction cosines of the line passing through the two points `( 2, 4, 5)`and `(1, 2, 3)`. |
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Answer» (x_1,y_1,z_1)=A and (x_2,y_2,z_2)=B dintance between A and B AB=`sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1))` =`sqrt(9+4+64)` =`sqrt77` `(l,m,n)=((x_2-x_1)/(AB),(y_2-y_1)/(AB),(z_2-z_1)/(AB))` after putting the values =`(3/sqrt77,2/sqrt77,8/sqrt77)` |
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| 12. |
Find the angle between the lines whose direction cosines are given by the equations `3l + m + 5n = 0` and `6mn - 2nl + 5lm = 0` |
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Answer» Direction Cosines of the two lines are given by the equations `3l+m+5n=0` and `6mn-2l+5lm=0` From first equation we get, `m=-5n-3l` Put it in second equation we get, `-30n^2-45nl-15l^2=0` `2n^2+3nl+l^2=0` `2n^2+2nl+nl+l^2=0` `(2n+l)(n+l)=0` If `l=-2n`, then `m=n` (by putting it in equation 1) and if `l=-n` we get `m=-2n` `costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2)` We get, `theta=cos^(-1)(-1/6)` |
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| 13. |
Direction cosines of the line passing through the points `A(-4, 2, 3) and B(1, 3, -2)` areA. `pm (1)/(sqrt(51)), pm (5)/(sqrt(51)), pm (1)/(sqrt(51))`B. `pm (5)/(sqrt(51)), pm (1)/(sqrt(51)), pm (-5)/(sqrt(51))`C. `pm 5, pm 1, pm 5`D. `pm sqrt(51), pm sqrt(51), pm sqrt(51)` |
| Answer» Correct Answer - B | |
| 14. |
Direction cosines of the line passing through the points `A(-4, 2, 3) and B(1, 3, -2)` areA. `-5, 1, -5`B. `5, 1, -5`C. `5, -1, 5`D. `5, 1, 5` |
| Answer» Correct Answer - B | |
| 15. |
Find the angle between theline whose direction cosines are givenby `l+m+n=0a n d2l^2+2m^2-n^2-0.` |
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Answer» `l^(2)+m^(2)+n^(2)=1" "`(i) `" "l+m+n=0" "`(ii) `" "2l^(2)+2m^(2) -n^(2)=0" "`(iii) `" "2(l-n^(2))=n^(2)or 3n^(2)=2 or n=pmsqrt(2//3)" "`(iv) `" "2(l^(2)+m^(2))=n^(2)=(-(l+m))^(2)or l=m" "`(v) `" "l+m=pmsqrt(2//3) or 2l=pmsqrt(2//3)` `" "l=pm1//sqrt(6), m=pm1//sqrt(6)` Direction cosines are `" "((1)/(sqrt(6)),(1)/(sqrt(6)), sqrt((2)/(3)))and ((1)/(sqrt(6)),(1)/(sqrt(6)),-sqrt((2)/(3)))` or `" "(-(1)/(sqrt(6)),-(1)/(sqrt(6)), sqrt((2)/(3)))and (-(1)/(sqrt(6)), -(1)/(sqrt(6)), -sqrt((2)/(3)))` The angle between these lines in both the cases is `cos^(-1)(-(1)/(3))`. |
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| 16. |
The angle between the lines whose direction cosines `l`, m, n satisfy the equations `5l+m+3n=0` and `5mn-2nl+6lm=0` isA. `cos^(-1)((1)/(sqrt(6)))`B. `cos^(-1)((1)/(sqrt(3)))`C. `cos^(-1)((1)/(6))`D. `cos^(-1)((1)/(3))` |
| Answer» Correct Answer - C | |
| 17. |
Find the direction cosines of the two lines whichare connected by th relations. `l-5m+3n=0a n d7l^2+5m^2-3n^2=0`A. `pm(-1)/(sqrt(6)), pm (1)/(sqrt(6)), pm (2)/(sqrt(6))`B. `pm (1)/(sqrt(6)), pm (1)/(sqrt(6)), pm (-2)/(sqrt(6))`C. `pm (2)/(sqrt(6)), pm (2)/(sqrt(6)), pm (-1)/(sqrt(6))`D. `pm (2)/(sqrt(6)), pm (2)/(sqrt(6)), pm (1)/(sqrt(6))` |
| Answer» Correct Answer - A | |
| 18. |
The direction cosines l, m and n of two lines are connected by the relations `l+m+n=0 and lm=0`, then the angle between the lines isA. `0^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - C | |
| 19. |
The direction cosines of aline satisfy the relations `lambda(l+m)=na n dm n+n l+l m=0.`The value of `lambda,`for which the two lines are perpendicular toeach other, isa. `1`b. `2`c. `1//2`d. noneof theseA. 1B. 2C. `1//2`D. none of these |
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Answer» Correct Answer - b Eliminating `n`, we get `" "lamda(l+m)^(2)+lm=0` `rArr" "(lamdal^(2))/(m^(2))+ (2lamda+1)(l)/(m) + lamda=0` `rArr" "(l_1l_2)/(m_(1)m_(2))=1" "` (product of roots `(l_1)/(m_1) and (l_2)/(m_2)` ) where `l_1//m_1 and l_2//m_2` are the roots of this equation, further eliminating m, we get `" "lamdal^(2)-ln-n^(2)=0` `rArr" "(l_(1)l_(2))/(n_(1)n_(2))=-(1)/(lamda)` Since the lines with direction cosines `(l_1, m_1, n_1) and (l_2, m_2, n_2)` are perpendicular, we have `" "l_1l_2+ m_1m_2 + n_1n_2 = 0` or `" "1+1-lamda=0` or `" "lamda=2` |
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| 20. |
The intercepts made on the axesby the plane the which bisects the line joining the points `(1,2,3)`and `(-3,4,5)`at right angles area. `(-9/2,9,9)`b. `(9/2,9,9)`c. `(9,-9/2,9)`d. `(9,9/2,9,)`A. `(-(9)/(2),9,9)`B. `((9)/(2),9,9)`C. `(9,-(9)/(2),9)`D. `(9,(9)/(2),9)` |
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Answer» Correct Answer - a Direction ratios of the joining points `P(1, 2, 3) and Q(-3, 4, 5)` are -4, 2, 2 which are the direction ratio of the normal to the plane. Then, equation of plane is `-4x+2y+2z=k`. Also this plane passes through the midpoint of `PQ(-1, 3, 4)`. Thus, `" "-4(-1)+ 2(3)+ 2(4)=k` or `" "k=18` Hence, equation of plane is `2x-y-z=-9` Then, intercepts are `(-9//2), 9 and 9` |
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| 21. |
Find the equation of a plane which cuts the interscepts `4,3` and `-2` units on `x,y` and z-axes respectively. |
| Answer» Correct Answer - `3x+4y-6z-12` | |
| 22. |
A line with positive direction cosines passes through the point P(2, – 1, 2) and makes equal angles with thecoordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQequals |
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Answer» Correct Answer - `sqrt3` The direction cosines of the line are `1//sqrt3, 1//sqrt3,,1//sqrt3`. Any point on the line at a distance `t` from `p(2, -1, 2)` is `(2+(t)/(sqrt3), -1+ (t)/(sqrt3), 2+ (t)/(sqrt3))`, which lies on `2x+y+z-9=0` `rArr" "t=sqrt3` |
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| 23. |
Find the equation of the plane through (3,4,-1) which is parallel to the plane `vecr.(2hati-3hatj+5hatk)+7=0` |
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Answer» The equation of any plane which is parallel to `vecr*(2hati-3hatj+5hatk)+7=0` is `" "vecr*(2hati-3hatj+5hatk)+lamda=0" "` (i) or `" "2x-3y+5z+lamda=0` Further (i) will pass through (3, 4, -1) if `(2)(3)+(-3)(4)+5(-1)+lamda=0 or -11+lamda=0 or lamda=0 or lamda=11` Thus, equation of the required plane is `vecr*(2hati-3hatj+5hatk)+11=0`. |
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| 24. |
Convert the equation of plane `2x-4y+3z = 24` into intercept from and find the intercepts cuts from the axes. |
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Answer» Equation of plane `2x-4y+3z = 24` `rArr (2x)/(24)-(4y)/(24) + (3z)/(24) = 1` `rArr (x)/(12) + (y)/(-6) + (z)/(8) = 1`. Which is the intercept from of plane. Intercepts on axes are `a = 12, b = - 6, c = 8` |
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| 25. |
Find the equation of the plane which cuts the intercepts of length `3,-4` and `2 `units on the axes respectively. |
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Answer» Here, `a = 3, b = - 4, c = 2` Equation of plane `x/a + y/b + z/c = 1` `rArr x/3 - y/4 + z/2 = 1` `rArr 4x-3y+6z = 12`. |
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| 26. |
A variable plane passesthrough a fixed point `(a ,b ,c)`and cuts the coordinateaxes at points `A ,B ,a n dCdot`Show that eh locus of thecentre of the sphere `O A B Ci s a/x+b/y+c/z=2.` |
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Answer» Let `(alpha, beta, gamma)` be any point on the locus. Then according to the given condition, `(alpha, beta, gamma)` is the centre of the sphere through the origin. Therefore, its equation is given by `" "(x-alpha)^(2)+(y-beta)^(2)+(z-gamma)^(2)=(0-alpha)^(2)+(0-beta)^(2)+(0-gamma)^(2)` `" "x^(2)+y^(2)+z^(2)-2alphax-2betay-2gammaz=0` To obtain its point of intersection with the x-axis, we put y=0 and z=0, so that `" "x^(2)-2alphax=0 or x(x-2alpha)=0 or x=0, 2alpha` Thus, the plane meets the x-axis at O (0, 0, 0) and `A(2alpha, 0, 0)`. Similarly, it meets the y-axis at `O(0, 0, 0) and B(0, 2beta, 0)`, and the z-axis at `O(0, 0, 0) and C(0, 0, 2gamma)`. The equation of the plane through `A, B and C` is `" "(x)/(2alpha)+(y)/(2beta)+(z)/(2gamma) =1" "`(intercept form) Since it passes through `(a, b, c)`, we get `" "(a)/(2alpha)+(b)/(2beta)+(c)/(2gamma)=1 or (a)/(alpha)+(b)/(beta)+(c)/(gamma)=2` Hence, locus of `(alpha, beta, gamma)` is `(a)/(x)+(b)/(y)+(c)/(z)=2` |
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| 27. |
A point `P`moves on a plane `x/a+y/b+z/c=1.`A plane through `P`and perpendicular to `O P`meets the coordinate axes at `A , Ba n d Cdot`If the planes through `A ,Ba n dC`parallel to the planes `x=0,y=0a n dz=0,`respectively, intersect at `Q ,`find the locus of `Qdot` |
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Answer» Correct Answer - `(1)/(ax)+ (1)/(by)+(1)/(cz)= (1)/(x^(2))+(1)/(y^(2))+ (1)/(z^(2))` The given plane is `(x)/(a)+(y)/(b)+(z)/(c)=1" "(i)` Let P(h,k,l) be the point on the plane. Then `(h)/(a)+(k)/(b)+(l)/(c)=1" "(ii)` `impliesOPsqrt(h^(2)+k^(2)+l^(2))` Direction cosines of OP are `(h)/sqrt(h^(2)+k^(2)+l^(2)),(k)/sqrt(h^(2)+k^(2)+l^(2))` and `(h)/sqrt(h^(2)+k^(2)+l^(2))` The equation of the plane through P and normal to OP is `(hx)/sqrt(h^(2)+k^(2)+l^(2))+(ky)/sqrt(h^(2)+k^(2)+l^(2))+(lz)/sqrt(h^(2)+k^(2)+l^(2))` `=sqrt(h^(2)+k^(2)+l^(2))` or `hx+ky+lz=h^(2)+k^(2)+l^(2)` Therefore, `A-=((h^(2)+k^(2)+l^(2))/(h),0,0)`, `B-=(0,(h^(2)+k^(2)+l^(2))/(k),0)` and `C-=(0,0,(h^(2)+k^(2)+l^(2))/(l))` If Q `(alpha,beta,gamma)`, then `alpha=(h^(2)+k^(2)+l^(2))/(h),beta=(h^(2)+k^(2)+l^(2))/(k)` and `gamma=(h^(2)+k^(2)+l^(2))/(l)" "(iii)` Now, `(1)/(a^(2))+(1)/(beta^(2))+(1)/(gamma^(2))` `=(h^(2)+k^(2)+l^(2))/((h^(2)+k^(2)+l^(2))^(2))=(1)/(h^(2)+k^(2)+l^(2))" "(iv)` From (iii), `h=(h^(2)+k^(2)+l^(2))/(alpha)or(h)/(a)=(h^(2)+k^(2)+l^(2))/(aalpha)` Similarly, `(k)/(b)=(h^(2)+k^(2)+l^(2))/(b beta)and(1)/(c)=(h^(2)+k^(2)+l^(2))/(cgamma)` `(h^(2)+k^(2)+l^(2))/(aalpha)+(h^(2)+k^(2)+l^(1))/(b beta)+(h^(2)+k^(2)+l^(2))/(cgamma)` `=(h)/(a)+(k)/(b)+(l)/(c)=1" "["from"(ii)]` or `(1)/(aalpha)+(1)/(b beta)+(1)/(cgamma)=(1)/(h^(2)+k^(2)+l^(2))=(1)/(alpha)+(1)/(beta^(2))+(1)/(gamma^(2))` [from (iv)] The required equation of locus is `(1)/(ax)+(1)/(by)+(1)/(cz)+(1)/(x^(2))+(1)/(y^(2))+(1)/(z^(2)` |
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| 28. |
If a plane meets the equations axes at `A ,Ba n dC`such that the centroid ofthe triangle is `(1,2,4),`then find the equation of the plane. |
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Answer» Let the plane meet the coordinates axes at `A(a, 0, 0), B(0, b, 0), and C(0, 0, c).` Then, `" "a=3, b=6, c=12`. Hence, the equation of required plane is `(x)/(3)+(y)/(6)+(z)/(12)=1 or 4x+2y+z=12` |
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| 29. |
Find the equation of a plane passes through the point `(4,4,1)` and the ratio of intercepts cuts on axes from this plane is `2 : 1 : 1`. |
| Answer» Correct Answer - `x+2y+2z=14` | |
| 30. |
Find th equation of the plane which is at distance of 8 units from origin and the perpendicular vector from origin to this plane is `(2hati+hatj-2hatk)`. |
| Answer» Correct Answer - `vecr.(2hati+hatj-2hatk)=24` | |
| 31. |
A sphere of constant radius`k ,`passes through the originand meets the axes at `A ,Ba n d Cdot`Prove that the centroid oftriangle `A B C`lies on the sphere `9(x^2+y^2+z^2)=4k^2dot` |
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Answer» Let the equation of any sphere passing through the origin and having radius k be `" "x^(2)+y^(2)+z^(2)+2ux+2vy+2wz=0" "` (i) As the radius of the sphere is k, we get `" "u^(2)+v^(2)+w^(2)=k^(2)" "`(ii) Note that (i) meets the x-axis at `O(0, 0, 0) and A(-2u, 0, 0), ` y-axis at `O(0, 0, 0) and B(0, -2v, 0),` and z-axis at `O(0, 0, 0) and C(0, 0, -2w)`. Let the centroid of the triangle `ABC` be `(alpha, beta, gamma)`. Then `" "alpha =-(2u)/(3), beta=-(2v)/(3), gamma =- (2w)/(3) rArr" " u =- (3alpha)/(2), v =- (3beta)/(2), w =- (3v)/(2)` Putting this in (ii), we get `" "((-3)/(2)alpha)^(2)+((-3)/(2)beta)^(2) + ((-3)/(2)gamma )^(2) = k^(2)` or `" "alpha^(2)+beta^(2)+gamma^(2)=(4)/(9) k^(2)` This shows that the centroid of triangle `ABC` lies on `x^(2)+y^(2)+z^(2)=(4)/(9) k^(2)`. |
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| 32. |
A plane meets the coordinate axes at `A ,Ba n dC`respectively such that the centroid of triangle `A B C`is `(1,-2,3)dot`Find the equation of the plane.A. `alphax+betay+gammaz=1`B. `alpha^(2)x+beta^(2)y+gamma^(2)z=3`C. `(x)/(alpha)+(y)/(beta)+(z)/(gamma) = 1`D. `(x)/(alpha)+(y)/(beta)+(z)/(gamma) = 3`. |
| Answer» Correct Answer - D | |
| 33. |
A plane meets the coordinate axes at `A ,Ba n dC`respectively such that the centroid of triangle `A B C`is `(1,-2,3)dot`Find the equation of the plane. |
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Answer» `(x/3,y/3,z/3)=(1,-2,3)` `x/3=1,y/3=-2,z/3=3` `x=3,y=6,z=9` `A(3,0,0),B(0,-6,0),C(0,0,9)` `AB=(-3,-9,0)` `BC=(0,6,9)` `AB*BC=|[i,j,k],[-3,-6,0],[0,6,9]|` `hati(-54)-hatj(-27)+hatk(-18)=0` `6hati-3hatj+2hatk=0` `6(x-x_1)+(-3)(y-y_!)+2(z-z_1)=0` `x_1=3,y_1=0,z_1=0` `6(x-3)-3y+2z=0` `6x-3y+2z-18=0` This is the equation of plane. |
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| 34. |
A variable plane is at a constant distance `p`from the origin and meets the coordinate axes in `A , B , C`. Show that the locus of the centroid of the tehrahedron `O A B Ci sx^(-2)+y^(-2)+z^(-2)=16p^(-2)dot` |
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Answer» Let the equation of variable plane be `x/a+y/b+z/c = 1 "……"(1)` Which meets the axes at `A(a,0,0), B(0,b,0)` and `C(0,0,c)` respectively. Let the centroid of `DeltaABC` is `(alpha,beta,gamma)` `alpha = (a+0+0)/(3),beta = (0+b+0)/(3), gamma = (0+0+c)/(3)` `rArr alpha = 3alpha , b = 3beta, c = 3gamma "......"(2)` Now the length of perpendicular from `(0,0,0)` to plane `(1) = 3p` `rArr |(0+0+0-1)/(sqrt(1/(a^(2))+(1)/(b^(2))+(1)/(c^(2))))|=3p` `rArr (1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))=(1)/(9p^(2))` `rArr 1/(9alpha^(2)) + (1)/(9beta^(2)) + (1)/(9gamma^(2)) = (1)/(9p^(2))` [From eq. (2)] `rArr alpha^(-2) + beta^(-2) + gamma^(-2) = p^(-2)` `:.` Locus of `(alpha, beta, gamma)` `x^(-2)+y^(-2)+z^(-2) = p^(-2)`. |
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| 35. |
Find the perpendicular distasnce of the point (1,0,0) from the lines (x-1)/2=(y+1)/(-3)=(z+10)/8`A. `sqrt(6)` unitB. `2sqrt(6)` unitC. `3sqrt(6)` unitD. None of these |
| Answer» Correct Answer - B | |
| 36. |
If A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2) are the vertices of `Delta` ABC, then the direction cosines of side AB areA. `(2)/(sqrt(17)), (2)/(sqrt(17)), (-3)/(sqrt(17))`B. `(2)/(sqrt(17)),(3)/(sqrt(17)),(2)/(sqrt(17))`C. `(4)/(sqrt(42)), (5)/(sqrt(42)), (-1)/(sqrt(42))`D. `(-4)/(sqrt(17)), (-5)/(sqrt(17)), (1)/(sqrt(17))` |
| Answer» Correct Answer - A | |
| 37. |
Show that the points A(2,3,5),B(-4,7,-7),C(-2,1,-10) and D(4,-3,2) are the vertices of a rectangle. |
| Answer» Show that AB=CD,BC=DA and AC=BD | |
| 38. |
Find the direction cosines of the sides of the triangle whose vertices are `(3, 5, 4)`, `( 1, 1, 2)`and `( 5, 5, 2)`. |
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Answer» Let `A(3,5,-4),B(-1,1,2) and C(-5,-5,-2)` are the given vertices of triangle `ABC`. Direction cosines of AB can be given as, `((x_2-x_1)/(AB),(y_2-y_1)/(AB),(z_2-z_1)/(AB)).` Here,` AB = sqrt((-1-3)^2+(1-5)^2+(2-(-4))^2) = sqrt(16+16+36) = sqrt68` `:.` Direction cosines of `AB = (-4/sqrt68,-4/sqrt68,6/sqrt68)` Similarly, we can find the direction cosines of `AC` and `BC`. |
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| 39. |
Show that the points `A(-1, 4, -3), B(3, 2, -5), C(-3, 8, -5)a n d D(-3, 2, 1)`are coplanar. |
| Answer» Show that DA=DB=DC | |
| 40. |
Show that the points `A(1,3,4), B(-1,6, 10), C(-7,4,7)a n d D(-5,1,1)`are have vertices of a rhombus. |
| Answer» Show that PQ=QR=RS=SP and PR `ne` QS | |
| 41. |
If A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2) are the vertices of `Delta` ABC, then the direction cosines of side AC areA. `(2)/(sqrt(17)), (2)/(sqrt(17)), (-3)/(sqrt(17))`B. `(2)/(sqrt(17)),(3)/(sqrt(17)),(2)/(sqrt(17))`C. `(4)/(sqrt(42)), (5)/(sqrt(42)), (-1)/(sqrt(42))`D. `(-4)/(sqrt(17)), (-5)/(sqrt(17)), (1)/(sqrt(17))` |
| Answer» Correct Answer - C | |
| 42. |
prove that point `A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a rhombusA. squareB. rhombusC. rectangleD. parallelogram |
| Answer» Correct Answer - B | |
| 43. |
The points (5, -4, 2), (4, -3, 1), (7, -6, 4) and (8, -7, 5) are the vertices of aA. squareB. rectangleC. rhombusD. parallelogram |
| Answer» Correct Answer - D | |
| 44. |
Find the equation of theplane containing the lines `(x-5)/4=(y-7)/4=(z+3)/(-5)a n d(x-8)/7=(y-4)/1=(z-5)/3dot` |
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Answer» The required plane is `" "|{:(x-5,,y-7,,z+3),(4,,4,,-5),(7,,1,,3):}|=0` or `" "17(x-5)-(12+35)(y-7)+(4-28)(z+3)=0` or `" "17x-47y-24z+172=0` |
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| 45. |
Find the distance betweenthe line `(x+1)/(-3)=(y-3)/2=(z-2)/1`and the plane `x+y+z+3=0.` |
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Answer» The line is along the vector `veca= - 3hati+2hatj+hatk ` and plane is normal to the vector `vecb= hati+hatk+hatk`. Since `veca*vecb = 0` , the line is parallel to the plane. Hence, the distance between the line and the plane is the distance of point `(-1, 3, 2)` from the plane, `" "(|-1+3+2+3|)/(sqrt(1+1+1))= (7)/(sqrt(3))` |
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| 46. |
Find the distance of thepoint `(-1,-5,-10)`from the point ofintersection of the line `(x-2)/3=(y+1)/4=(z-2)/(12)`and plane `x-y+z=5.` |
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Answer» Any point on the line `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)= lamda` is `(3lamda+2, 4lamda-1, 12lamda+2)`. This lies on `x-y+z=5`. If `3lamda+2-4lamda+1+12lamda+2=5 or lamda=0`, then the point is `(2, -1, 2)`. Its distance from `(-1, -5, -10)` is `" "sqrt((2+1)^(2)+(-1+5)^(2)+(2+10)^(2))` `" "=sqrt(9+16+144)=13` |
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| 47. |
Find the equation of theplane passing through the points `(1,0,-1)a n d(3,2,2)`and parallel to the line `x-1=(1-y)/2=(z-2)/3dot` |
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Answer» The equation of any plane through the point `(1, 0, -1)` is `" "A(x-1)+B(y-0)+C(z+1)=0" "`(i) Since it passes through the point `(3, 2, 2)`, we get `" "2A+2B+3C=0" "`(ii) Since plane (i) is parallel to the line `" "(x-1)/(1)=(y-1)/(-2)=(z-2)/(3)` we have `1A+(-2)B+3C=0" "` (iii) From (i) and (iii), `" "A:B:C= 4:-1 : -2` Substituting these values in (i), we get `" "4(x-1)-1(y-0)-2(z+1)=0` or `" "4x-y-2z-6=0` |
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| 48. |
Find the equation of the plane passing through the point `( 1, 3, 2)`and perpendicular to each of the planes `x + 2y + 3z = 5`and `3x + 3y + z = 0`. |
| Answer» Correct Answer - `2x-4y+3z=8` | |
| 49. |
Find the angle between the planes `3x+y+2z=1` and `2x-y+z+3 = 0`. |
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Answer» Here `{:(a_(1)=3,b_(1)=1,c_(1)=2),(a_(2)=2,b_(2)=-1,c_(2)=1):}` If the angle between the planes is `theta`, then `costheta = (a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1^(2))+b_(1^(2))+c_(1^(2)))sqrt(a_(2^(2))+b_(2^(2))+c_(2^(2))))` `=(6-1+2)/(sqrt(9+1+4)sqrt(4+1+1))` `=(7)/(sqrt(14)sqrt(6)) = (7)/(2sqrt(21))` `rArr theta = cos^(-1)(7/(2sqrt(21)))` |
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| 50. |
Find the vector equation of the following plane in product form `vecr = ( hati-hatk)+lambda(hati+2hatj)+mu(hati+3hatj-hatk)`. |
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Answer» Comparing with `vecr = veca+lambdavecb+muvecc` `veca=hati-hatk,vecb=hati+2hatj,vecc=hati+3hatj-hatk` `vecn=vecbxxvecc=|{:(hati,hatj,hatk),(1,2,0),(1,3,-1):}|` `=hati(-2-0)-hatj(-1-0)+hatk(3-2)` `=-2hati+hatj+hatk` Equation of plane `vecr.vecn=veca.vecn` `rArr vecr.(-2hati+hatj+hatk) = (hati-hatk).(-2hati+hatj+hatk)` `= -2-1` `= - 3` `rArr vecr.(-2hati+hatj+hatk) = - 3` |
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