JEE Mathematics Three Dimensional Geometry MCQs Set 26

Answer: (a) \( \frac{x+3}3=\frac{y-5}1=\frac{z-2}{-5} \)

Explanation:
1. The direction \( (3,1,-5) \) satisfies \( 2(3)-1-5=0 \), so the line is parallel to the plane and its image has the same direction.
2. Reflect \( (1,3,4) \): the normal line \( (2k+1,3-k,4+k) \) has midpoint on the plane when \( k=-2 \), so the image is \( (-3,5,2) \).
3. The image line is \( \frac{x+3}3=\frac{y-5}1=\frac{z-2}{-5} \). Answer: (a)

Teacher's Note:
1. Check whether the line is parallel to the plane.
2. Reflect a point of the line.

Answer: (c) \( \frac72 \)

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

Teacher's Note:
1. Make the normals equal.

Answer: (c) exactly two values

Explanation:
1. Coplanarity: \( \begin{vmatrix}2-1&3-4&4-5\\1&1&-k\\k&2&1\end{vmatrix}=0 \), i.e. \( \begin{vmatrix}1&-1&-1\\1&1&-k\\k&2&1\end{vmatrix}=0 \).
2. \( 1(1+2k)+1(1+k^2)-1(2-k)=k^2+3k=0 \).
3. \( k=0 \) or \( k=-3 \), exactly two values. Answer: (c)

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

Answer: (c) \( \frac92 \)

Explanation:
1. Points \( (2p+1,3p-1,4p+1) \) and \( (q+3,2q+k,q) \) coincide: \( 2p+1=q+3 \), \( 3p-1=2q+k \), \( 4p+1=q \).
2. From the first and third: \( p=-\frac32 \), \( q=-5 \).
3. Then \( 3\left(-\frac32\right)-1=2(-5)+k\Rightarrow k=\frac92 \). Answer: (c)

Teacher's Note:
1. Equate the coordinates.
2. Solve for \( p,q \) first.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The midpoint of \( AB \) is \( (1,3,5) \), which satisfies \( \frac11=\frac{3-1}2=\frac{5-2}3=1 \), so the Reason is true.
2. \( \overrightarrow{AB}=(0,6,-4) \) and the line direction is \( (1,2,3) \): \( 0+12-12=0 \), so they are perpendicular.
3. A line that bisects the segment and is perpendicular to it makes the points mirror images, so the Reason explains the Assertion. Answer: (a)

Teacher's Note:
1. Check the midpoint.
2. Check perpendicularity.

Answer: (d) \( \sqrt{53} \)

Explanation:
1. Point on the line \( Q(2\lambda,3\lambda+2,4\lambda+3) \). \( \overrightarrow{PQ}=(2\lambda-3,3\lambda+3,4\lambda-8) \) is perpendicular to \( (2,3,4) \).
2. \( 2(2\lambda-3)+3(3\lambda+3)+4(4\lambda-8)=29\lambda-29=0 \), so \( \lambda=1 \) and \( Q=(2,5,7) \).
3. \( PQ=\sqrt{1+36+16}=\sqrt{53} \). Answer: (d)

Teacher's Note:
1. Find the foot of the perpendicular.

Answer: (c) \( 60^\circ \)

Explanation:
1. \( \cos^245^\circ+\cos^2120^\circ+\cos^2\theta=1\Rightarrow\frac12+\frac14+\cos^2\theta=1 \).
2. \( \cos^2\theta=\frac14 \), so \( \cos\theta=\pm\frac12 \); the acute angle is \( 60^\circ \). Answer: (c)

Teacher's Note:
1. Use \( l^2+m^2+n^2=1 \).

Answer: (b) \( (-6,7) \)

Explanation:
1. The direction \( (3,-5,2) \) is perpendicular to the normal \( (1,3,-\alpha) \): \( 3-15-2\alpha=0\Rightarrow\alpha=-6 \).
2. The point \( (2,1,-2) \) lies on the plane: \( 2+3+6(-2)+\beta=0\Rightarrow\beta=7 \).
3. \( (\alpha,\beta)=(-6,7) \). Answer: (b)

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

Answer: (d) \( a=6,b=4 \)

Explanation:
1. Points on the line: \( (5-2t,\,1+t(b-1),\,a+t(1-a)) \). At \( x=0 \), \( t=\frac52 \).
2. \( 1+\frac52(b-1)=\frac{17}2\Rightarrow b=4 \) and \( a+\frac52(1-a)=-\frac{13}2\Rightarrow a=6 \). Answer: (d)

Teacher's Note:
1. Parametrize the line and set \( x=0 \).

Answer: (b) -5

Explanation:
1. Intersecting lines are coplanar: \( \begin{vmatrix}k&2&3\\3&k&2\\1&1&-2\end{vmatrix}=0 \) (the third row is the vector joining the two given points, up to sign).
2. \( k(-2k-2)-2(-6-2)+3(3-k)=-2k^2-5k+25=0 \), i.e. \( 2k^2+5k-25=0 \).
3. \( k=\frac52 \) or \( k=-5 \); the integer value is \( -5 \). Answer: (b)

Teacher's Note:
1. Use the coplanarity condition.
2. Solve the quadratic.

Answer: (a) \( \frac1{\sqrt3} \)

Explanation:
1. Direction ratios \( (l,m,n) \) satisfy \( 2l+3m+n=0 \) and \( l+3m+2n=0 \), giving \( (l,m,n)\propto(3,-3,3)\propto(1,-1,1) \).
2. Direction cosines: \( \pm\frac1{\sqrt3} \), so \( \cos\alpha=\frac1{\sqrt3} \) for the positive direction. Answer: (a)

Teacher's Note:
1. Use the cross product of the normals.

Answer: (a) \( (4,9,-3) \)

Explanation:
1. The centre is \( (3,6,1) \) and it is the midpoint of the diameter.
2. Other end \( (\alpha,\beta,\gamma) \): \( \frac{\alpha+2}2=3,\ \frac{\beta+3}2=6,\ \frac{\gamma+5}2=1 \), so \( (4,9,-3) \). Answer: (a)

Teacher's Note:
1. Find the centre.
2. Use the midpoint formula.

Answer: (d) \( \left(\frac95,-\frac{13}5,4\right) \)

Explanation:
1. \( \frac{x+1}1=\frac{y-3}{-2}=\frac{z-4}0=\frac{-2(-1-6)}{1+4}=\frac{14}5 \).
2. \( x=\frac{14}5-1=\frac95 \), \( y=3-\frac{28}5=-\frac{13}5 \) and \( z=4 \). Answer: (d)

Teacher's Note:
1. Use the image formula.

Answer: (b) -2

Explanation:
1. The centres are \( (-3,4,1) \) and \( (5,-2,1) \), with midpoint \( (1,1,1) \).
2. \( 2a-3a+4a+6=0\Rightarrow3a=-6 \), so \( a=-2 \). Answer: (b)

Teacher's Note:
1. Find both centres.

Answer: (d) \( \frac53 \)

Explanation:
1. \( \sin\theta=\frac{|2-2+2\sqrt\lambda|}{3\sqrt{5+\lambda}}=\frac13 \), so \( 2\sqrt\lambda=\sqrt{5+\lambda} \).
2. \( 4\lambda=5+\lambda\Rightarrow\lambda=\frac53 \). Answer: (d)

Teacher's Note:
1. Use the line-plane angle formula.

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

Explanation:
1. \( 2x=3y=-z \) gives direction \( (3,2,-6) \), and \( 6x=-y=-4z \) gives \( (2,-12,-3) \).
2. Dot product \( =6-24+18=0 \), so the angle is \( 90^\circ \). Answer: (c)

Teacher's Note:
1. Convert to direction ratios.

Answer: (c) 1

Explanation:
1. Sphere centre \( \left(\frac12,0,-\frac12\right) \), radius \( \sqrt{\frac14+\frac14+2}=\frac{\sqrt{10}}2 \).
2. Distance of the centre from the plane \( =\frac{|\frac12+0+\frac12-4|}{\sqrt6}=\frac3{\sqrt6} \).
3. Circle radius \( =\sqrt{\frac{10}4-\frac96}=1 \). Answer: (c)

Teacher's Note:
1. Find the centre and radius of the sphere.
2. Use Pythagoras.

Answer: (c) \( \frac35 \)

Explanation:
1. \( 2\cos^2\theta+\cos^2\beta=1\Rightarrow2\cos^2\theta=\sin^2\beta=3\sin^2\theta \).
2. \( 2\cos^2\theta=3(1-\cos^2\theta)\Rightarrow\cos^2\theta=\frac35 \). Answer: (c)

Teacher's Note:
1. Use \( l^2+m^2+n^2=1 \).

Answer: (b) \( (3a,2a,3a),(a,a,a) \)

Explanation:
1. Points: \( F(k,k-a,k) \) on the first line and \( E(2\lambda-a,\lambda,\lambda) \) on the second.
2. \( \overrightarrow{EF}\parallel(2,1,2) \): \( \frac{k-2\lambda+a}2=\frac{k-\lambda-a}1=\frac{k-\lambda}2 \), giving \( k=3a \) and \( \lambda=a \).
3. So \( F=(3a,2a,3a) \) and \( E=(a,a,a) \). Answer: (b)

Teacher's Note:
1. Take a general point on each line.
2. Impose the direction ratios.

Answer: (a) -2

Explanation:
1. Directions \( (1,-\lambda,\lambda) \) and \( (1,2,-2) \); the joining vector is \( (1,-4,-1) \).
2. Coplanarity: \( \begin{vmatrix}1&-4&-1\\1&-\lambda&\lambda\\1&2&-2\end{vmatrix}=-10-5\lambda=0 \).
3. \( \lambda=-2 \). Answer: (a)

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

Answer: (d) \( 2x-y-z=1 \)

Explanation:
1. The common plane is the difference of the two equations: \( 10x-5y-5z-5=0 \).
2. This is \( 2x-y-z=1 \). Answer: (d)

Teacher's Note:
1. Subtract the sphere equations.

Answer: (c) 3

Explanation:
1. The sphere has centre \( (-1,1,2) \) and radius \( \sqrt{1+1+4+19}=5 \).
2. Distance of the centre from the plane \( =\frac{|-1+2+4+7|}3=4 \).
3. Circle radius \( =\sqrt{25-16}=3 \). Answer: (c)

Teacher's Note:
1. Find the centre and radius.
2. Use Pythagoras.

Answer: (c) 13

Explanation:
1. The sphere has centre \( (-2,1,3) \) and radius \( \sqrt{4+1+9+155}=13 \).
2. Distance of the centre from the plane \( =\frac{|-24+4+9-327|}{13}=\frac{338}{13}=26 \).
3. Shortest distance \( =26-13=13 \). Answer: (c)

Teacher's Note:
1. Find the distance from the centre.
2. Subtract the radius.

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

Explanation:
1. The plane is \( \frac xa+\frac yb+\frac zc=1 \) in one system and \( \frac x{a'}+\frac y{b'}+\frac z{c'}=1 \) in the other.
2. The perpendicular distance from the common origin is the same: \( \frac1{\sqrt{\frac1{a^2}+\frac1{b^2}+\frac1{c^2}}}=\frac1{\sqrt{\frac1{a'^2}+\frac1{b'^2}+\frac1{c'^2}}} \).
3. Hence \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}-\frac1{a'^2}-\frac1{b'^2}-\frac1{c'^2}=0 \). Answer: (d)

Teacher's Note:
1. Use the intercept form in both systems.
2. Equate the distance from the origin.

Answer: (a) 7 units

Explanation:
1. The edges are \( 5-2=3 \), \( 9-3=6 \) and \( 7-5=2 \).
2. Diagonal \( =\sqrt{9+36+4}=7 \) units. Answer: (a)

Teacher's Note:
1. Find the edge lengths.

Answer: (b) \( al+bm+cn=0 \)

Explanation:
1. The plane passes through \( (x_1,y_1,z_1) \) and contains the line, so its normal \( (a,b,c) \) is perpendicular to the direction \( (l,m,n) \).
2. Hence \( al+bm+cn=0 \). Answer: (b)

Teacher's Note:
1. Relate the normal to the line direction.

Answer: (b) \( (1,3,4) \)

Explanation:
1. The sphere has centre \( (0,1,2) \). The line through it normal to the plane is \( (\lambda,1+2\lambda,2+2\lambda) \).
2. On the plane: \( \lambda+2+4\lambda+4+4\lambda=15\Rightarrow9\lambda=9 \), so \( \lambda=1 \).
3. The circle's centre is \( (1,3,4) \). Answer: (b)

Teacher's Note:
1. Drop a perpendicular from the sphere's centre.