JEE Mathematics Three Dimensional Geometry MCQs Set 25

Answer: (c) \( x-y-z=0 \)

Explanation:
1. The family of planes is \( (2x-y-4)+\lambda(y+2z-4)=0 \).
2. Through \( (1,1,0) \): \( -3-3\lambda=0\Rightarrow\lambda=-1 \).
3. The plane is \( 2x-2y-2z=0 \), i.e. \( x-y-z=0 \). Answer: (c)

Teacher's Note:
1. Use the family of planes.
2. Substitute the given point.

Answer: (a) greater than 3 but less than 4

Explanation:
1. A point on the line is \( A(10r-3,-7r+2,r) \). For the foot, \( \overrightarrow{PA}\cdot(10,-7,1)=0 \): \( 10(10r-5)-7(3-7r)+(r-4)=0\Rightarrow150r=75 \), \( r=\frac12 \).
2. Foot \( \left(2,-\frac32,\frac12\right) \) and distance \( =\sqrt{0+\frac14+\frac{49}4}=\frac5{\sqrt2}\approx3.54 \). Answer: (a)

Teacher's Note:
1. Parametrize the line.
2. Use perpendicularity to find the foot.

Answer: (b) \( \frac92 \)

Explanation:
1. \( P=(1+2r,-1+3r,2+4r) \). Substituting: \( 20r=10 \), so \( r=\frac12 \) and \( P=\left(2,\frac12,4\right) \).
2. \( OP=\sqrt{4+\frac14+16}=\frac92 \). Answer: (b)

Teacher's Note:
1. Parametrize the line.
2. Substitute in the plane.

Answer: (a) \( (\sqrt2,1,4) \)

Explanation:
1. Let the plane be \( ax+by+cz=d \). The two points give \( -b=d \) and \( c=d \), so the plane is \( ax-dy+dz=d \).
2. \( \cos\frac\pi4=\frac{2|d|}{\sqrt{a^2+2d^2}\sqrt2}\Rightarrow a^2+2d^2=4d^2\Rightarrow a=\pm\sqrt2d \). Planes: \( \pm\sqrt2x-y+z=1 \).
3. Only \( (\sqrt2,1,4) \) lies on \( -\sqrt2x-y+z=1 \) (\( -2-1+4=1 \)); the other options satisfy neither plane. Answer: (a)

Teacher's Note:
1. Find the plane in terms of \( a,d \).
2. Use the angle formula and test the options.

Answer: (c) \( \frac{11}{\sqrt5} \)

Explanation:
1. Family: \( (1+2\lambda)x+(1+3\lambda)y+(1+\lambda)z+5\lambda-6=0 \). Perpendicular to the \( XY \)-plane means the \( z \)-coefficient is 0: \( \lambda=-1 \).
2. The plane is \( -x-2y-11=0 \), i.e. \( x+2y+11=0 \). Distance of \( (0,0,256) \) \( =\frac{11}{\sqrt5} \). Answer: (c)

Teacher's Note:
1. Make the coefficient of \( z \) zero.

Answer: (a) \( \frac{\sqrt{91}}2 \)

Explanation:
1. The image of \( Q \) in the plane is \( P \): with \( 3(0)-(-1)+4(-3)-2=-13 \), \( P=Q-2\cdot\frac{-13}{26}(3,-1,4)=(3,-2,1) \).
2. \( \overrightarrow{PQ}=(-3,1,-4) \) and \( \overrightarrow{PR}=(0,1,-3) \); cross product \( =(1,-9,-3) \) of magnitude \( \sqrt{91} \).
3. Area \( =\frac{\sqrt{91}}2 \). Answer: (a)

Teacher's Note:
1. Reflect \( Q \) in the plane.
2. Use half the cross product.

Answer: (a) 13

Explanation:
1. First plane written as \( 2x-y+2z+\frac\lambda2=0 \): distance \( \frac{|\lambda/2-3|}3=\frac13 \), so \( \lambda=8 \) or \( 4 \).
2. Third: \( \frac{|\mu-3|}3=\frac23 \), so \( \mu=5 \) or \( 1 \).
3. Maximum \( \lambda+\mu=8+5=13 \). Answer: (a)

Teacher's Note:
1. Make the normals identical.
2. Use the parallel-plane distance formula.

Answer: (c) \( (2,0,1) \)

Explanation:
1. Point on the line \( (2r+1,-1-r,r) \). The foot on \( x+y+z=3 \) is \( \left(2+\frac{4r}3,-\frac{5r}3,1+\frac r3\right) \).
2. It lies on \( x-y+z=3 \): \( 2+\frac{4r}3+\frac{5r}3+1+\frac r3=3\Rightarrow r=0 \).
3. \( Q=(2,0,1) \). Answer: (c)

Teacher's Note:
1. Find the foot in terms of \( r \).
2. Impose the second plane.

Answer: (d) \( 2\sqrt{14} \)

Explanation:
1. Points on the line: \( (3r+2,2r-1,-r+1) \). With the first plane: \( 13r+13=0 \), \( r=-1 \), so \( P(-1,-3,2) \).
2. With the second plane: \( 7r=7 \), \( r=1 \), so \( Q(5,1,0) \).
3. \( PQ=\sqrt{36+16+4}=\sqrt{56}=2\sqrt{14} \). Answer: (d)

Teacher's Note:
1. Parametrize the line.
2. Find \( r \) for each plane.

Answer: (c) \( \sqrt3 \)

Explanation:
1. Normal \( =(1,2,-1)\times(-1,1,-2)=(-3,3,3) \), so the plane through \( (1,1,0) \) is \( x-y-z=0 \).
2. Distance of \( (2,1,4) \): \( \frac{|2-1-4|}{\sqrt3}=\sqrt3 \). Answer: (c)

Teacher's Note:
1. Find the normal.
2. Use the point-to-plane formula.

Answer: (c) \( (3,2,1) \)

Explanation:
1. Family: \( (1+2\lambda)x+(1+3\lambda)y+(1-\lambda)z+(4\lambda-1)=0 \). Parallel to the \( Y \)-axis: \( 1+3\lambda=0 \), \( \lambda=-\frac13 \).
2. The plane is \( x+4z-7=0 \); only \( (3,2,1) \) satisfies it. Answer: (c)

Teacher's Note:
1. Make the \( y \)-coefficient zero.

Answer: (c) \( aa'+c+c'=0 \)

Explanation:
1. The first line has direction \( (a,1,c) \) and the second has direction \( (a',c',1) \).
2. Perpendicular: \( aa'+c'+c=0 \). Answer: (c)

Teacher's Note:
1. Write each direction vector.

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

Explanation:
1. Normal of the second plane \( \mathbf n_1=(3,4,2)\times(4,2,3)=(8,-1,-10) \).
2. The required plane contains the origin and the directions \( (2,3,4) \) and \( \mathbf n_1 \), so its normal is \( (2,3,4)\times(8,-1,-10)=(-26,52,-26)\parallel(1,-2,1) \).
3. The plane is \( x-2y+z=0 \). Answer: (d)

Teacher's Note:
1. Find the normal of the plane of the two lines.
2. The required normal is perpendicular to both the line and that normal.

Answer: (c) \( (1,1,1) \)

Explanation:
1. Normal \( =(3,-1,2)\times(1,2,3)=(-7,-7,7)\parallel(1,1,-1) \).
2. The plane is \( (x-4)+(y+1)-(z-2)=0 \), i.e. \( x+y-z-1=0 \).
3. Only \( (1,1,1) \) satisfies it. Answer: (c)

Teacher's Note:
1. Take the cross product of the directions.
2. Test each point.

Answer: (d) \( (4,1,-2) \)

Explanation:
1. The midpoint is \( (0,2,5) \) and the normal is along \( (6,10,2)\parallel(3,5,1) \).
2. The plane is \( 3x+5(y-2)+(z-5)=0 \), i.e. \( 3x+5y+z=15 \).
3. Only \( (4,1,-2) \) satisfies it (\( 12+5-2=15 \)). Answer: (d)

Teacher's Note:
1. Use the midpoint and direction of the segment.
2. Test the points.

Answer: (b) \( \sqrt2,1,-1 \)

Explanation:
1. Let the plane be \( ax+by+cz+b=0 \) with \( c+b=0 \) (it passes through both points).
2. \( \cos\frac\pi4=\frac{|b-c|}{\sqrt{a^2+b^2+c^2}\sqrt2}\Rightarrow a^2+b^2+c^2=(b-c)^2\Rightarrow a^2=-2bc=2b^2 \).
3. So \( a=\pm\sqrt2b \) and \( c=-b \); the ratios \( \sqrt2:1:-1 \) fit. Answer: (b)

Teacher's Note:
1. Use the two points to relate \( b \) and \( c \).
2. Apply the angle formula.

Answer: (c) \( (2,0,-2) \)

Explanation:
1. The required plane contains the line (direction \( \mathbf b=(2,-1,3) \)) and the normal of the given plane \( \mathbf n=(2,3,-1) \).
2. Its normal is \( \mathbf b\times\mathbf n=(-8,8,8)\parallel(1,-1,-1) \), and it passes through \( (3,-2,1) \): \( x-y-z=4 \).
3. Only \( (2,0,-2) \) satisfies it. Answer: (c)

Teacher's Note:
1. Find the plane through the line and the normal direction.
2. Test the points.

Answer: (a) \( (2,-4,-7) \)

Explanation:
1. Points: \( (a+3,3a-1,-a+6) \) and \( (7b-5,-6b+2,4b+3) \). Equating gives \( a-7b=-8 \), \( a+2b=1 \) and \( a+4b=3 \), so \( b=1 \), \( a=-1 \).
2. \( R=(2,-4,7) \). Its reflection in the \( xy \)-plane is \( (2,-4,-7) \). Answer: (a)

Teacher's Note:
1. Equate the two parametrizations.
2. Reflect by changing the sign of \( z \).

Answer: (c) \( \cos^{-1}\left(\frac{19}{35}\right) \)

Explanation:
1. Normal of \( OPQ \): \( \overrightarrow{OP}\times\overrightarrow{OQ}=(1,2,1)\times(2,1,3)=(5,-1,-3) \).
2. Normal of \( PQR \): \( \overrightarrow{PQ}\times\overrightarrow{PR}=(1,-1,2)\times(-2,-1,1)=(1,-5,-3) \).
3. \( \cos\theta=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35} \). Answer: (c)

Teacher's Note:
1. Find the normals of both faces.
2. Angle between faces is the angle between normals.

Answer: (a) \( \{\sqrt3,-\sqrt3\} \)

Explanation:
1. The four points are coplanar, so the determinant of the vectors from \( (-1,-1,1) \) to the other three points vanishes:
2. \( \begin{vmatrix}1-\lambda^2&2&0\\2&1-\lambda^2&0\\2&2&-\lambda^2-1\end{vmatrix}=0\Rightarrow(-1-\lambda^2)[(1-\lambda^2)^2-4]=0 \).
3. \( (1+\lambda^2)^2(3-\lambda^2)=0 \) gives \( \lambda^2=3 \), so \( \lambda=\pm\sqrt3 \). Answer: (a)

Teacher's Note:
1. Use the coplanarity determinant.

Answer: (a) \( \sqrt{\frac53} \)

Explanation:
1. \( \cos\theta=\frac{2\sqrt2}3\Rightarrow\sin\theta=\frac13 \).
2. \( \sin\theta=\frac{|2-2+2k|}{3\sqrt{5+k^2}}=\frac13\Rightarrow2|k|=\sqrt{5+k^2} \).
3. \( 4k^2=5+k^2\Rightarrow k^2=\frac53 \), so \( k=\pm\sqrt{\frac53} \). Answer: (a)

Teacher's Note:
1. Convert the cosine to a sine.
2. Use the line-plane angle formula.

Answer: (d) \( \sqrt{\frac23} \)

Explanation:
1. \( \overrightarrow{AB}=(-1,0,-1) \) with length \( \sqrt2 \), and the normal is \( (1,1,1) \).
2. \( \sin\theta=\frac{|-2|}{\sqrt2\sqrt3}=\sqrt{\frac23} \), so \( \cos\theta=\frac1{\sqrt3} \).
3. Projection \( =AB\cos\theta=\frac{\sqrt2}{\sqrt3}=\sqrt{\frac23} \). Answer: (d)

Teacher's Note:
1. Find the angle between the segment and the normal.
2. Projection length is \( AB\cos\theta \).

Answer: (b) \( \frac{10}{\sqrt{83}} \) units

Explanation:
1. Normal \( =(1,-2,3)\times(2,-1,-1)=(5,7,3) \).
2. Plane through \( (1,-1,-1) \): \( 5(x-1)+7(y+1)+3(z+1)=0 \), i.e. \( 5x+7y+3z+5=0 \).
3. Distance of \( (1,3,-7) \): \( \frac{|5+21-21+5|}{\sqrt{83}}=\frac{10}{\sqrt{83}} \). Answer: (b)

Teacher's Note:
1. Take the cross product of the directions.
2. Use the distance formula.

Answer: (b) \( 10\sqrt3 \)

Explanation:
1. Point on the line through \( (1,-5,9) \) with direction \( (1,1,1) \): \( (\lambda+1,\lambda-5,\lambda+9) \).
2. On the plane: \( (\lambda+1)-(\lambda-5)+\lambda+9=5\Rightarrow\lambda=-10 \), giving \( (-9,-15,-1) \).
3. Distance \( =\sqrt{10^2+10^2+10^2}=10\sqrt3 \). Answer: (b)

Teacher's Note:
1. Parametrize the line.
2. Find where it meets the plane.

Answer: (d) 2

Explanation:
1. The line is perpendicular to the normal: \( 2l-m-3=0 \).
2. The point \( (3,-2,-4) \) lies on the plane: \( 3l-2m+4=9\Rightarrow3l-2m=5 \).
3. Solving: \( l=1 \), \( m=-1 \), so \( l^2+m^2=2 \). Answer: (d)

Teacher's Note:
1. Use the direction condition.
2. Use the point condition.

Answer: (c) \( x+3y+6z=7 \)

Explanation:
1. Family: \( (2+\lambda)x+(\lambda-5)y+(4\lambda+1)z-3-5\lambda=0 \).
2. Parallel to \( x+3y+6z=1 \): \( \frac{2+\lambda}1=\frac{\lambda-5}3=\frac{4\lambda+1}6 \), giving \( \lambda=-\frac{11}2 \).
3. The plane becomes \( x+3y+6z-7=0 \). Answer: (c)

Teacher's Note:
1. Write the family of planes.
2. Equate the ratios of coefficients.