JEE Mathematics Three Dimensional Geometry MCQs Set 20

Answer: (d) 2 and 4

Explanation:
1. For a line of intersection the homogeneous system needs non-trivial solutions: \( \begin{vmatrix}k&4&1\\4&k&2\\2&2&1\end{vmatrix}=0 \).
2. Expanding: \( k(k-4)-4(0)+(8-2k)=k^2-6k+8=0 \), so \( k=2,4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 2

Explanation:
1. \( \hat k\times(\hat k\times\vec\alpha)=\hat k(\hat k\cdot\vec\alpha)-\vec\alpha=\gamma\hat k-\vec\alpha=-\alpha\hat i-\beta\hat j \), so \( \alpha=\beta=0 \).
2. Then \( \gamma=2 \) from the plane.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

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

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac72 \)

Explanation:
1. Rewrite the second plane as \( 2x+y+2z+\frac52=0 \).
2. Distance \( =\frac{|-8-\frac52|}{\sqrt{4+1+4}}=\frac{21/2}{3}=\frac72 \).

Teacher's Note:
1. Make the coefficients of x, y, z identical first.
2. Use |d1 - d2| / |n|.

Answer: (b) \( 2\sqrt2 \)

Explanation:
1. Normal \( =(2,-2,1)\times(1,-1,2)=(-3,-3,0) \), so the plane is \( x+y=c \).
2. Through \( (1,-2,1) \): \( c=-1 \), i.e. \( x+y+1=0 \).
3. Distance from \( (1,2,2) \): \( \frac{1+2+1}{\sqrt2}=2\sqrt2 \).

Teacher's Note:
1. The normal of the required plane is the cross product of the two given normals.
2. Apply the point-to-plane distance formula.

Answer: (a) (ii) and (iii) only

Explanation:
1. \( PQ=\frac{|1+2+3-3|}{\sqrt3}=\sqrt3 \).
2. The foot is \( P-t(1,1,1) \) with \( t=\frac{6-3}{3}=1 \): \( Q=(0,1,2) \).

Teacher's Note:
1. Distance from a point to a plane, then move along the normal.
2. Check that Q satisfies x + y + z = 3.

Answer: (c) \( l=m=n=-\frac23 \)

Explanation:
1. Conditions: \( l+2m+2=0 \), \( 2l+2+n=0 \), \( 2+m+2n=0 \).
2. Solving gives \( l=m=n=-\frac23 \) (check: \( -\frac23-\frac43+2=0 \)).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (d) m + 2n = -2

Explanation:
1. Perpendicular lines satisfy \( 1\cdot2+m\cdot1+2\cdot n=0 \), i.e. \( m+2n=-2 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (b) 6

Explanation:
1. The plane of the lines has normal \( (2,3,4)\times(3,4,5)=(-1,2,-1)\parallel(1,-2,1) \) and passes through \( (1,2,3) \): \( x-2y+z=0 \).
2. The given plane is parallel to it, so \( A=1 \) and the distance is \( \frac{|d|}{\sqrt6}=\sqrt6 \), giving \( |d|=6 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 7

Explanation:
1. From the second equation \( z=3x \), the first gives \( y=0 \), and the third holds. Points \( (x,0,3x) \).
2. Need \( 10x^2\le100 \): \( x=-3,\dots,3 \), i.e. 7 points.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \sqrt3 \)

Explanation:
1. Direction cosines \( \frac1{\sqrt3}(1,1,1) \). Point \( (2+s,-1+s,2+s) \) at distance \( s\sqrt3 \).
2. Plane: \( 2(2+s)+(-1+s)+(2+s)=9\Rightarrow4s=4 \), so \( s=1 \) and \( PQ=\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac{17}{5\sqrt3} \)

Explanation:
1. Points \( A(-1,-2,-1) \) and \( B(2,-2,3) \): \( \vec{AB}=(3,0,4) \).
2. Common normal \( (-1,-7,5) \): \( \frac{|\vec{AB}\cdot n|}{|n|}=\frac{|-3+0+20|}{\sqrt{75}}=\frac{17}{5\sqrt3} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( n=-2(l+1)=-\frac{m+2}{2} \)

Explanation:
1. A normal to the plane \( AOB \) is \( OA\times OB=(4-m,\ 1-2l,\ lm-2) \), which must be proportional to \( (2,1,n) \).
2. So \( 4-m=2(1-2l) \), giving \( m=4l+2 \), and \( n=\frac{lm-2}{1-2l}=-2(l+1) \) which also equals \( -\frac{m+2}2 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (c) \( \frac{1}{5\sqrt3}(-\hat i-7\hat j+5\hat k) \)

Explanation:
1. \( (3,1,2)\times(1,2,3)=(1\cdot3-2\cdot2,\ 2\cdot1-3\cdot3,\ 3\cdot2-1\cdot1)=(-1,-7,5) \).
2. Its magnitude is \( \sqrt{75}=5\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( x-2y+z=0 \)

Explanation:
1. Normal of the second plane: \( (3,4,2)\times(4,2,3)=(8,-1,-10) \).
2. Our plane contains direction \( (2,3,4) \) and is perpendicular to \( (8,-1,-10) \), so its normal is \( (2,3,4)\times(8,-1,-10)=(-26,52,-26)\parallel(1,-2,1) \). Through the origin: \( x-2y+z=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac49\sqrt{29} \)

Explanation:
1. The moved point is \( (1+2t,\ 2+3t,\ 3+4t) \). On the plane: \( 6+9t=2 \), so \( t=-\frac49 \).
2. Distance \( =|t|\sqrt{4+9+16}=\frac49\sqrt{29} \).

Teacher's Note:
1. Move along the direction vector with parameter t.
2. Distance equals |t| times the length of the direction vector.

Answer: (c) Statement 1 is false and Statement 2 is true

Explanation:
1. Directions: \( n_2\times n_3=(0,-4,-4) \), \( n_3\times n_1=(0,2,2) \), \( n_1\times n_2=(0,2,2) \): all parallel to \( (0,1,1) \), so Statement 1 is false.
2. Adding \( P_1,P_2 \) gives \( x=0 \) and \( z-y=1 \); \( P_3 \) with \( x=0 \) gives \( z-y=\frac23 \): inconsistent, so no common point (Statement 2 true).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \left(\frac83,\frac43,-\frac73\right) \)

Explanation:
1. \( \frac{|1-4-2-\alpha|}{3}=5 \) with \( \alpha>0 \) gives \( \alpha=10 \).
2. At \( P \) the plane expression is \( -5 \), so move \( \frac{15}{9}=\frac53 \) along \( (1,2,-2) \): \( \left(1+\frac53,-2+\frac{10}3,1-\frac{10}3\right)=\left(\frac83,\frac43,-\frac73\right) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( -\frac12 \)

Explanation:
1. A normal to the plane is \( \vec{OP}\times\vec{OQ}=(2\lambda,\,-2\lambda,\,-6) \).
2. This is parallel to \( (1,-1,6) \), so \( (2\lambda,-2\lambda,-6)=k(1,-1,6) \) gives \( k=-1 \).
3. Hence \( 2\lambda=-1 \), \( \lambda=-\frac12 \).

Teacher's Note:
1. The normal of a plane through the origin is the cross product of two position vectors in it.
2. Perpendicular to the plane means parallel to the normal.

Answer: (b) \( \frac14 \)

Explanation:
1. \( Q=(1-3\mu,\ -1+\mu,\ 2+5\mu) \), so \( \vec{PQ}=(-2-3\mu,\ -3+\mu,\ -4+5\mu) \).
2. Perpendicular to the normal \( (1,-4,3) \): \( -2-3\mu+12-4\mu-12+15\mu=0 \), so \( 8\mu=2 \), \( \mu=\frac14 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 3a

Explanation:
1. Points: \( P=(t,\ t-a,\ t) \) and \( Q=(s-a,\ \tfrac s2,\ \tfrac s2) \).
2. Parallel to \( (2,1,2) \): subtracting the y and z components gives \( a=-k \); then \( s=2a \) and \( t=3a \).
3. So \( P=(3a,2a,3a) \), \( Q=(a,a,a) \), \( PQ=\sqrt{4a^2+a^2+4a^2}=3a \).

Teacher's Note:
1. Parametrise both lines and impose the direction ratios.
2. Use the distance formula at the end.

Answer: (c) \( lmn+9=2(l+m+n) \)

Explanation:
1. Coplanar through the origin means \( \begin{vmatrix}l&2&1\\1&m&2\\2&1&n\end{vmatrix}=0 \).
2. Expanding: \( lmn-2l-2n+8+1-2m=0 \), i.e. \( lmn+9=2(l+m+n) \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (d) \( \frac{\sqrt5}3 \)

Explanation:
1. The line has direction \( (1,2,2) \) (the normal) and passes through \( P(0,1,0) \).
2. Distance of the origin: \( \frac{|\vec{OP}\times\vec d|}{|\vec d|}=\frac{|(2,0,-1)|}{3}=\frac{\sqrt5}3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 1

Explanation:
1. The normal of the plane is \( (3,5,7) \). The plane contains the second line, so its direction \( (1,-2,k) \) is perpendicular to the normal.
2. So \( 3-10+7k=0 \) and \( k=1 \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (c) Statement 1 is false and Statement 2 is true

Explanation:
1. The direction is \( (3,-6,-2)\times(2,1,-2)=(14,2,15) \), so Statement 2 is true.
2. The point \( (3,1,0) \) does not lie on the first plane: \( 9-6-0=3\ne15 \). So the given line does not lie on the planes and Statement 1 is false.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.