JEE Mathematics Three Dimensional Geometry MCQs Set 14

Answer: (a) \( \sqrt3 \)

Explanation:
1. The direction cosines are \( \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right) \); a point is \( (2+t,-1+t,2+t) \) (t is the common step in each coordinate).
2. Substituting: \( 2(2+t)+(-1+t)+(2+t)=9 \) gives t = 1.
3. The point is (3, 0, 3) and \( PQ=\sqrt{1+1+1}=\sqrt3 \).

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At (2, 1, 5): \( 4+2-10-1=-5 \); at (3, 4, 3): \( 6+8-6-1=7 \).
2. The signs differ, so the points are on opposite sides, and A is true.
3. R is exactly the criterion and explains A.

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) \( \frac32 \)

Explanation:
1. The intercepts are a = 1, b = 2, c = -1.
2. Area \( =\frac12\sqrt{(bc)^2+(ca)^2+(ab)^2}=\frac12\sqrt{4+1+4} \).
3. The area is \( \frac32 \).

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) in no point

Explanation:
1. A point of the line is (t, 2t, 3t).
2. Substituting: \( 2t-8t+6t=0\neq3 \) for every t.
3. So the line is parallel to the plane and does not meet it.

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) \( x^2+y^2+z^2-2cz=0 \)

Explanation:
1. Let the lines pass through \( (q,0,0),(0,r,0),(0,0,c) \). Mutual perpendicularity gives \( S-qx-ry=0,\ S-qx-cz=0,\ S-ry-cz=0 \) with \( S=x^2+y^2+z^2 \).
2. Subtracting: \( qx=cz=ry \).
3. Substituting back: \( S-2cz=0 \).

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) \( \pi-4 \)

Explanation:
1. Coplanarity: \( \begin{vmatrix}2&3&\lambda\\3&2&3\\-1&1&1\end{vmatrix}=0 \), i.e. \( -20+5\lambda=0 \) and \( \lambda=4 \).
2. Since 4 is outside \( [-\frac\pi2,\frac\pi2] \), \( \sin^{-1}(\sin4)=\pi-4 \).

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) \( \frac3{\sqrt{11}},\frac{-1}{\sqrt{11}},\frac1{\sqrt{11}} \)

Explanation:
1. The line of intersection has direction \( (2,1,-5)\times(4,-3,7)\propto(4,17,5) \).
2. The line of greatest slope is perpendicular to it within the plane: \( (2,1,-5)\times(4,17,5)=(90,-30,30)\propto(3,-1,1) \).
3. Divide by \( \sqrt{11} \).

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) 8 : 1 : 16

Explanation:
1. \( p_1=\frac{4}{\sqrt{29}} \); the second plane is \( 2x-3y+4z+\frac32=0 \) so \( p_2=\frac{1/2}{\sqrt{29}} \); \( p_3=\frac8{\sqrt{29}} \).
2. The ratio is 4 : 0.5 : 8, i.e. 8 : 1 : 16.

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) A is false, but R is true.

Explanation:
1. \( \sin\theta=\frac{|2-3+2|}{\sqrt{17}\sqrt3}=\frac1{\sqrt{51}} \), which is positive and not \( -\frac2{\sqrt{51}} \), so A is false.
2. R is the standard definition and is true.

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) A is false, but R is true.

Explanation:
1. PQ = (-6, -6, 6) and PR = (-2, -2, 2) are proportional, so P, Q, R are collinear.
2. Infinitely many planes contain a line, so A is false.
3. R is true: three collinear points do not fix a 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: (c) 4

Explanation:
1. \( L_1 \) and \( L_2 \) meet at the origin, and \( L_3 \) passes through \( (1,1,1) \) on \( L_1 \).
2. Let the third vertex be \( (t,2t,3t) \). Area \( =\frac12|(1,1,1)\times(t,2t,3t)|=\frac12|(t,-2t,t)|=\frac{\sqrt6|t|}2 \).
3. Setting this equal to \( \sqrt6 \) gives \( t=2 \), so \( \beta=2t=4 \).

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: (b) \( \frac2{\sqrt6} \)

Explanation:
1. Direction \( (1,2,3)\times(3,2,1)=(-4,8,-4)\propto(1,-2,1) \).
2. \( \cos\phi=\frac{|-2|}{\sqrt6}=\frac2{\sqrt6} \).

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) -1 : 3

Explanation:
1. \( 1=\frac{2m_2+4m_1}{m_1+m_2} \) gives \( -3m_1=m_2 \).
2. So \( m_1:m_2=-1:3 \) (an external division).

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) (-1, 0, 0)

Explanation:
1. The plane is \( 3x+y-1+\lambda(z-4)=0 \).
2. The given line has direction \( (1,1,1)\times(0,1,2)=(1,-2,1) \), so \( 3-2+\lambda=0 \) and \( \lambda=-1 \).
3. The plane is \( 3x+y-z+3=0 \), which meets the x-axis at \( (-1,0,0) \).

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: (c) \( 3x+y=5 \)

Explanation:
1. The bisector with the minus combination contains the origin.
2. The other one is \( \frac{2x-y+z-2}{\sqrt6}+\frac{x+2y-z-3}{\sqrt6}=0 \).
3. This gives \( 3x+y-5=0 \).

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) 4

Explanation:
1. \( L=\frac{2abc}{\sqrt{a^2b^2+b^2c^2+c^2a^2}} \) from the earlier calculation.
2. \( L^2\cdot\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}=4 \).

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) \( \vec a+2(d-\vec a\cdot\hat n)\hat n \)

Explanation:
1. The image is twice as far from A as the foot is, along the same direction.
2. So image \( =\vec a+2(d-\vec a\cdot\hat n)\hat n \).

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) \( \pi \)

Explanation:
1. Distance of the end from the plane is \( \frac{|1-0-2+4|}{3}=1 \).
2. The rod sweeps a cone of height 1 and slant height 2, so the radius is \( \sqrt{4-1}=\sqrt3 \).
3. Volume \( =\frac13\pi(3)(1)=\pi \).

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) \( |d-\vec a\cdot\hat n| \)

Explanation:
1. The distance of the plane from the origin is d.
2. The signed distance of A along the unit normal is \( \vec a\cdot\hat n \).
3. The separation is \( |d-\vec a\cdot\hat n| \).

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) \( \frac1{\sqrt6} \)

Explanation:
1. \( (1,2,3)\times(2,3,4)=(-1,2,-1) \) with length \( \sqrt6 \).
2. The joining vector is (1, 2, 4), so \( |(1,2,4)\cdot(-1,2,-1)|=1 \).
3. Distance \( \frac1{\sqrt6} \).

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) -1

Explanation:
1. Let \( Q=(k,k,1) \) with PQ perpendicular to direction (1, 1, 0); then \( (\lambda-k)+(\lambda-k)=0 \) so \( Q=(\lambda,\lambda,1) \). Similarly \( R=(0,0,-1) \).
2. \( \vec{PQ}=(0,0,1-\lambda) \) and \( \vec{PR}=(-\lambda,-\lambda,-1-\lambda) \), so the dot product is \( (1-\lambda)(-1-\lambda)=0 \).
3. Thus \( \lambda=\pm1 \); for \( \lambda=1 \) P coincides with Q, so \( \lambda=-1 \).

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: (d) 2

Explanation:
1. The plane through the line is \( (x+y+2z-3)+\lambda(2x+3y+4z-4)=0 \).
2. Parallel to the z-axis means the z-coefficient vanishes: \( 2+4\lambda=0 \), so \( \lambda=-\frac12 \).
3. The plane becomes \( y=-2 \), whose distance from the z-axis is 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) \( \frac{13}{\sqrt{74}} \)

Explanation:
1. The plane is \( 3x-4y+7z=6-12-7=-13 \), i.e. \( 3x-4y+7z+13=0 \).
2. Distance from the origin \( =\frac{13}{\sqrt{9+16+49}}=\frac{13}{\sqrt{74}} \).

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) \( \left(\frac{\alpha^2+\beta^2+\gamma^2}{\alpha},0,0\right) \)

Explanation:
1. The plane through P with normal OP is \( \alpha x+\beta y+\gamma z=\alpha^2+\beta^2+\gamma^2 \).
2. Putting y = z = 0 gives \( x=\frac{\alpha^2+\beta^2+\gamma^2}{\alpha} \).

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) \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}-\frac1{a'^2}-\frac1{b'^2}-\frac1{c'^2}=0 \)

Explanation:
1. The distance of the plane from the common origin is the same in both systems.
2. In each system this distance p satisfies \( \frac1{p^2}=\frac1{a^2}+\frac1{b^2}+\frac1{c^2} \).
3. Equating the two expressions gives the first option.

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.