JEE Mathematics Three Dimensional Geometry MCQs Set 16

Answer: (c) \( 2\sqrt2 \)

Explanation:
1. The normal is \( (2,-2,1)\times(1,-1,2)=(-3,-3,0)\propto(1,1,0) \).
2. The plane is \( x+y=1-2=-1 \), i.e. \( x+y+1=0 \).
3. Distance from (1, 2, 2) is \( \frac{4}{\sqrt2}=2\sqrt2 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( \frac xa+\frac yb+\frac zc=1 \)

Explanation:
1. The circumcentre of a triangle lies in the plane of the triangle.
2. The plane through A, B, C is \( \frac xa+\frac yb+\frac zc=1 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) 4

Explanation:
1. The second line has direction \( (3,-2,1)\times(2,3,4)=(-11,-10,13) \) and passes through \( \left(0,\frac{k+20}{11},\frac{2k-15}{11}\right) \).
2. Coplanarity with the first line (point (-4, -6, 1), direction (3, 5, -2)) gives \( k=4 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( \sqrt2 \)

Explanation:
1. \( x^2+y^2+z^2-xy-yz-zx=\frac12[(x-y)^2+(y-z)^2+(z-x)^2] \), so the surface is the line \( x=y=z \).
2. The nearest point to (1, 2, 3) is (2, 2, 2).
3. Distance \( =\sqrt{1+0+1}=\sqrt2 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) A is true, but R is false.

Explanation:
1. Adding the two planes of the line (lambda = 1) gives \( 5x+2z-8=0 \), so it contains the line; and \( 5(2)+0(-1)+2(-5)=0 \) shows it is perpendicular to the third plane, so A is true.
2. Points of the line are \( (1+t,-1+t,1+t) \); substituting in the plane gives \( 3+5t=5 \), i.e. \( t=\frac25 \), so the point is \( \left(\frac75,-\frac35,\frac75\right) \) and not (1, 1, 1).
3. R is false.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( \cos^{-1}\frac1{6\sqrt3} \)

Explanation:
1. Directions \( (-2,1,1) \) and \( (1,4,-1) \), lengths \( \sqrt6 \) and \( 3\sqrt2 \).
2. Dot product \( -2+4-1=1 \).
3. \( \cos\theta=\frac1{3\sqrt{12}}=\frac1{6\sqrt3} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) 3

Explanation:
1. Volume \( V=\frac16\cdot AB\cdot CD\cdot d\cdot\sin\theta \).
2. \( V=\frac16(1)(\sqrt3)(2)\frac{\sqrt3}{2}=\frac12 \).
3. Hence 6V = 3.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( \frac ax+\frac by+\frac cz=2 \)

Explanation:
1. The centre of the sphere through O, A, B, C is \( (x,y,z)=(p/2,q/2,r/2) \) where p, q, r are the intercepts.
2. The plane passes through (a, b, c): \( \frac ap+\frac bq+\frac cr=1 \).
3. Substituting p = 2x etc.: \( \frac ax+\frac by+\frac cz=2 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) \( \frac xf+\frac yg-\frac zh=0 \)

Explanation:
1. \( L=(0,g,h) \) and \( M=(f,0,h) \).
2. The normal is \( \vec{OL}\times\vec{OM}=(gh,hf,-fg) \).
3. Dividing by fgh: \( \frac xf+\frac yg-\frac zh=0 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) -3

Explanation:
1. Coplanarity: \( \begin{vmatrix}1&-1&-1\\1&1&-k\\k&2&1\end{vmatrix}=0 \).
2. Expanding gives \( k^2+3k=0 \), so k = 0 or k = -3.
3. The sum is -3.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) the ZX-plane

Explanation:
1. Direction \( (1,2,-1)\times(1,3,-1)=(1,0,1) \).
2. The y-component is zero, so the line is parallel to the ZX-plane.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( \frac{13}{\sqrt{75}} \)

Explanation:
1. The plane has normal (-1, -7, 5): \( -(x+1)-7(y+2)+5(z+1)=0 \).
2. At (1, 1, 1): \( -2-21+10=-13 \).
3. Distance \( \frac{13}{\sqrt{75}} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) (-1, 4, 3)

Explanation:
1. The plane is \( 2x-y+z+3=0 \); the value at the point is \( 2-3+4+3=6 \) and \( |n|^2=6 \).
2. Foot \( =(1,3,4)-1\cdot(2,-1,1)=(-1,4,3) \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( \frac1{\sqrt2} \)

Explanation:
1. Take \( O=(0,0,0),B=(4\sqrt2,0,0),C=(2\sqrt2,2\sqrt6,0),A=(0,0,2) \).
2. The directions are \( (2\sqrt2,0,-2) \) and \( (3\sqrt2,\sqrt6,0) \).
3. \( \cos\theta=\frac{12}{\sqrt{12}\sqrt{24}}=\frac1{\sqrt2} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( 90^\circ \)

Explanation:
1. Directions: (3, -2, 0) and \( \left(1,\frac32,2\right)\propto(2,3,4) \).
2. Dot product: \( 6-6+0=0 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) 7

Explanation:
1. The family through the line is \( 2x-2y-3+\lambda z=0 \).
2. At (3, 1, 1): \( 6-2-3+\lambda=0 \), so \( \lambda=-1 \).
3. The normals (2, -2, 1) and (2, -2, -1): \( \cos\theta=\frac{|4+4-1|}9=\frac79 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) \( \sin^2\theta+\sin^2\phi+\sin^2\psi+2\sin\theta\sin\phi\sin\psi=1 \)

Explanation:
1. A nonzero common solution needs \( \begin{vmatrix}1&-\sin\psi&-\sin\phi\\-\sin\psi&1&-\sin\theta\\-\sin\phi&-\sin\theta&1\end{vmatrix}=0 \).
2. Expanding: \( 1-\sin^2\theta-\sin^2\phi-\sin^2\psi-2\sin\theta\sin\phi\sin\psi=0 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( \frac x2=\frac{y-1}{-7}=\frac{z-2}{5} \)

Explanation:
1. A point of the line is \( (2\lambda-2,-\lambda-1,3\lambda) \); its plane value is \( 4\lambda-6 \).
2. The foot is \( P-\frac{4\lambda-6}{3}(1,1,1)=\left(\frac{2\lambda}3,\,1-\frac{7\lambda}3,\,2+\frac{5\lambda}3\right) \).
3. This traces the line through \( (0,1,2) \) with direction ratios (2, -7, 5).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( \cos^2\alpha=\cos^2\beta+\cos^2\gamma \)

Explanation:
1. The basic identity is \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \).
2. Then \( \sin^2\alpha=1-\cos^2\alpha=\cos^2\beta+\cos^2\gamma \) holds, and \( \sin^2\alpha+\sin^2\beta=2-(1-\cos^2\gamma)=1+\cos^2\gamma \) holds.
3. The relation \( \cos^2\alpha=\cos^2\beta+\cos^2\gamma \) would force the sum to be 2 cos^2 alpha = 1, which is not true in general.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) (5, 8, 15)

Explanation:
1. The foot is L = (3, 5, 9).
2. The image is \( 2L-P=(6-1,10-2,18-3)=(5,8,15) \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) (1, 1, 2)

Explanation:
1. Unit vectors are \( \hat{OA}=\frac{(-4,0,3)}5 \) and \( \hat{OB}=\frac{(14,2,-5)}{15} \).
2. The bisector direction is \( \hat{OA}+\hat{OB}=\frac{(2,2,4)}{15} \), proportional to (1, 1, 2).
3. Only (1, 1, 2) lies on this line through the origin.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( \cos^2 60^\circ+\cos^2 60^\circ+\cos^2\gamma=1 \) gives \( \cos^2\gamma=\frac12 \).
2. So \( \gamma=\frac\pi4 \) (or \( \frac{3\pi}{4} \)).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( \sqrt{\frac{17}3} \)

Explanation:
1. Points of L: \( (1+2t,t,-1-2t) \); \( (2t,t-1,-2-2t)\cdot(2,1,-2)=9t+3=0 \) so \( t=-\frac13 \).
2. \( P=\left(\frac13,-\frac13,-\frac13\right) \) and \( Q=2P-A=\left(-\frac13,-\frac53,-\frac53\right) \).
3. \( |OQ|^2=\frac{1+25+25}9=\frac{17}3 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) 4

Explanation:
1. \( \vec{AB}=(a-1,2-a,0) \) and \( \vec{AC}=(a-1,b-a,3-b) \).
2. For a different from 1 the first and second components force b = 2 and b = 3 together, which is impossible.
3. So a = 1; then \( (0,1,0)\parallel(0,b-1,3-b) \) gives b = 3 and a + b = 4.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.