JEE Mathematics Three Dimensional Geometry MCQs Set 15

Answer: (b) \( \frac{-\hat i-7\hat j+5\hat k}{5\sqrt3} \)

Explanation:
1. \( (3,1,2)\times(1,2,3)=(-1,-7,5) \).
2. Its length is \( \sqrt{75}=5\sqrt3 \).

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) 15

Explanation:
1. The YZ-plane is \( x=0 \); it divides the join in the ratio \( 3:2 \).
2. The point is \( \left(0,\frac{13}5,2\right) \).
3. \( x+5y+z=0+13+2=15 \).

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: (d) \( \left(-\infty,\frac{-1-\sqrt{73}}6\right)\cup\left(\frac{-1+\sqrt{73}}6,\infty\right) \)

Explanation:
1. At A the expression is 10 > 0.
2. At P: \( 3(\lambda^2+1)-\lambda+2(\lambda-1)-7=3\lambda^2+\lambda-6 \), which must be positive.
3. The roots are \( \frac{-1\pm\sqrt{73}}6 \), so the solution is outside the roots.

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) \( \frac14 \)

Explanation:
1. \( Q=(1-3\mu,-1+\mu,2+5\mu) \) so \( \vec{PQ}=(-2-3\mu,-3+\mu,-4+5\mu) \).
2. Parallel to the plane means the dot product with (1, -4, 3) is zero: \( -2-3\mu+12-4\mu-12+15\mu=-2+8\mu=0 \).
3. Hence \( \mu=\frac14 \).

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) \( 3x+y+z=25 \)

Explanation:
1. A point of the first line is \( (3+2a,2+3a,1+4a) \) and of the second \( (2+3b,3+2b,2+3b) \).
2. Equating gives a = b = 1, so the point is (5, 5, 5).
3. Checking: \( 3(5)+5+5=25 \).

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) -2

Explanation:
1. Plane A is the perpendicular bisector: \( 3x-3y+z-2=0 \).
2. \( \cos45^\circ=\frac{|3+3+c|}{\sqrt{19}\sqrt{2+c^2}}=\frac1{\sqrt2} \) gives \( 17c^2-24c-34=0 \).
3. The product of the roots 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: (d) 1

Explanation:
1. \( \frac{OD}{AD}=\frac{[OBC]}{[ABC]} \) and similarly for the others.
2. The three areas sum to the area of ABC, so the sum is 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: (c) \( L_1L_2 \) and \( L_2L_3 \) only

Explanation:
1. \( L_2 \) and \( L_3 \) have proportional directions (4, 2, 4) and (2, 1, 2), so they are parallel and coplanar.
2. For \( L_1,L_2 \) the triple product of the joining vector and the two directions is 0, so they intersect.
3. For \( L_1,L_3 \) it is 6, so they are skew.

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. With \( r^2=\alpha^2+\beta^2+\gamma^2 \), \( Q=\left(\frac{r^2}\alpha,\frac{r^2}\beta,\frac{r^2}\gamma\right) \), so \( \frac1{X^2}+\frac1{Y^2}+\frac1{Z^2}=\frac1{r^2} \).
2. The condition \( \frac\alpha a+\frac\beta b+\frac\gamma c=1 \) becomes \( \frac1{aX}+\frac1{bY}+\frac1{cZ}=\frac1{X^2}+\frac1{Y^2}+\frac1{Z^2} \).
3. At (1, 1, 1): \( \frac1a+\frac1b+\frac1c=3 \).

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) \( \sin\theta=\frac27 \)

Explanation:
1. \( \sin\theta=\frac{|2\cdot1|}{\sqrt{4+9+36}}=\frac27 \).

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) \( xyz=6k^3 \)

Explanation:
1. With intercepts a, b, c: \( \frac{abc}6=64k^3 \), so \( abc=384k^3 \).
2. The centroid of the tetrahedron OABC is \( (a/4,b/4,c/4) \).
3. \( xyz=\frac{abc}{64}=6k^3 \).

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: (c) \( l_1+l_2,\ m_1+m_2,\ n_1+n_2 \)

Explanation:
1. Take unit vectors OP and OQ along the two lines; the mid-point of PQ lies on the bisector.
2. Its coordinates are proportional to \( (l_1+l_2,m_1+m_2,n_1+n_2) \).

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: (b) 64

Explanation:
1. Points (0, b, 0) on L1 and (a, 0, 0) on L2, difference \( (a,-b,0) \); normal \( (bc,-ac,-ab) \).
2. \( d=\frac{2abc}{\sqrt{b^2c^2+c^2a^2+a^2b^2}}=\frac14 \).
3. Squaring: \( 64a^2b^2c^2=\sum a^2b^2 \), so \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=64 \).

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) \( \frac{\sqrt{29}}5 \)

Explanation:
1. Points along the line from the given point are \( (1+2t,-2+3t,3-4t) \).
2. On the plane: \( 1+2t+2-3t+3-4t=5 \), so \( t=\frac15 \).
3. Distance \( =\frac15\sqrt{4+9+16}=\frac{\sqrt{29}}5 \).

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: (d) \( 3\hat i+5\hat j+9\hat k \)

Explanation:
1. \( (a+\lambda b-P)\cdot b=(5+3\lambda,5+2\lambda,4-2\lambda)\cdot(3,2,-2)=17+17\lambda=0 \).
2. So \( \lambda=-1 \) and the foot is \( \vec a-\vec b=(3,5,9) \).

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) 10 : 11

Explanation:
1. At A: \( 3+4+10-7=10 \); at B: \( 0-6+2-7=-11 \).
2. The ratio is \( 10:11 \) (internal, since the signs differ).

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) P and S

Explanation:
1. \( OP^2=0+25+36=61 \), \( OQ^2=66 \), \( OR^2=62 \), \( OS^2=9+16+36=61 \).
2. P and S are equidistant and nearest.

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. \( \frac{1-\lambda}{\sqrt{(1+\lambda)^2+(1-\lambda)^2+4}}=\frac12 \).
2. Squaring: \( 4(1-\lambda)^2=2\lambda^2+6 \), i.e. \( \lambda^2-4\lambda-1=0 \).
3. The sum of the roots is 4 (the roots are \( 2\pm\sqrt5 \)).

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) \( \frac{20\pi}{\sqrt6} \)

Explanation:
1. The line is parallel to the normal (1, -1, 2) of both planes, so the cylinder axis is perpendicular to them.
2. The height is the distance between the planes: \( \frac{|3-(-2)|}{\sqrt6}=\frac{5}{\sqrt6} \).
3. Volume \( =\pi r^2h=\pi\cdot4\cdot\frac5{\sqrt6}=\frac{20\pi}{\sqrt6} \).

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: (c) \( \frac\pi4 \)

Explanation:
1. \( n_1\propto(1,0,0) \) and \( n_2\propto(1,-1,-1) \).
2. The line direction is \( n_1\times n_2=(0,1,-1) \).
3. \( \cos\theta=\frac{|1+2|}{\sqrt2\cdot3}=\frac1{\sqrt2} \), so the acute angle is \( \frac\pi4 \).

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) \( \frac\pi6 \)

Explanation:
1. The dot product is \( (a+c)\cos\theta+\sqrt2 b\sin\theta+\sqrt3(a-c) \), independent of theta only if \( a+c=0,\ b=0 \).
2. Then \( \cos\alpha=\frac{\sqrt3\cdot2|a|}{\sqrt8\cdot\sqrt2|a|}=\frac{\sqrt3}2 \).
3. So \( \alpha=\frac\pi6 \).

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. The direction of L is \( (2,1,4)\times(3,2,-2)=(-10,16,1) \), with \( |d|^2=357 \).
2. So \( Q=P\pm(-10,16,1) \): (-9, 20, 4) or (11, -12, 2).
3. The positive first coordinate gives (11, -12, 2) and the sum 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) \( 5x+4y+z-10=0 \)

Explanation:
1. The family is \( (x-y-z-2)+\lambda(x+2y+z-2)=0 \), with normal \( (1+\lambda,\,2\lambda-1,\,\lambda-1) \).
2. Perpendicular to the normal (1, -1, -1): \( (1+\lambda)-(2\lambda-1)-(\lambda-1)=3-2\lambda=0 \), so \( \lambda=\frac32 \).
3. The plane is \( 5x+4y+z-10=0 \).

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) A is false, but R is true.

Explanation:
1. The direction is \( (3,-6,-2)\times(2,1,-2)=(14,2,15) \), so R is true.
2. The point (3, 1, 0) gives \( 9-6-0=3\ne15 \), so it is not on the first plane; the correct line is \( x=3+14t,y=-1+2t,z=15t \). A 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.