JEE Mathematics Three Dimensional Geometry MCQs Set 22

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

Explanation:
1. Points on the line through the given point with direction \( (1,1,1) \): \( (1+t,-5+t,9+t) \).
2. Put into the plane: \( (1+t)-(-5+t)+(9+t)=15+t=5 \), so \( t=-10 \).
3. Distance \( =|t|\sqrt{1^2+1^2+1^2}=10\sqrt3 \).

Teacher's Note:
1. Distance along a direction equals |t| times the length of the direction vector.
2. The parameter t is found by substituting in the plane equation.

Answer: (a) \( 2\sqrt{42} \)

Explanation:
1. Points on the line through P with direction \( (1,4,5) \): \( (1+t,-2+4t,3+5t) \).
2. This meets the plane when \( 2(1+t)+3(-2+4t)-4(3+5t)+22=0 \), i.e. \( 6-6t=0 \), so \( t=1 \): the foot is at distance \( \sqrt{1+16+25}=\sqrt{42} \) from P.
3. The foot is the mid-point of P and Q, so \( PQ=2\sqrt{42} \).

Teacher's Note:
1. PQ is twice the oblique distance from P to the plane.
2. Find t at the intersection point first.

Answer: (d) \( \frac{10}{\sqrt{83}} \)

Explanation:
1. Direction ratios \( (1,-2,3) \) and \( (2,-1,-1) \) the normal is their cross product \( (2+3,6+1,-1+4)=(5,7,3) \).
2. Plane: \( 5(x-1)+7(y+1)+3(z+1)=0 \), i.e. \( 5x+7y+3z+5=0 \).
3. Distance \( =\frac{|5+21-21+5|}{\sqrt{25+49+9}}=\frac{10}{\sqrt{83}} \).

Teacher's Note:
1. A vector perpendicular to two directions is their cross product.
2. Use the point-to-plane distance formula.

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

Explanation:
1. The segment vector is \( (1,0,1) \) with length squared 2 the plane normal is \( (1,1,1) \).
2. Component along the normal \( =\frac{2}{\sqrt3} \), so its square is \( \frac43 \).
3. Projected length \( =\sqrt{2-\frac43}=\sqrt{\frac23} \).

Teacher's Note:
1. Subtract the normal component from the segment vector.
2. Use Pythagoras: projected^2 + normal component^2 = length^2.

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

Explanation:
1. The first line has direction \( (a,1,c) \) (taking y as parameter).
2. The second line has direction \( (a',c',1) \) (taking z as parameter).
3. Perpendicular means the dot product is zero: \( aa'+c'+c=0 \).

Teacher's Note:
1. Write each line in symmetric form to read off direction ratios.
2. Perpendicular lines have zero dot product of directions.

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

Explanation:
1. Normal to \( OPQ \): \( \vec{OP}\times\vec{OQ}=(5,-1,-3) \).
2. Normal to \( PQR \): \( \vec{PQ}\times\vec{PR}=(1,-1,1)\times(-2,-1,1)=(1,-5,-3) \).
3. Dot product \( 5+5+9=19 \) both magnitudes are \( \sqrt{35} \), so \( \cos\theta=\frac{19}{35} \).

Teacher's Note:
1. The angle between two faces is the angle between their normals.
2. Use cross products of edge vectors.

Answer: (b) \( \sqrt{34} \)

Explanation:
1. The line passes through \( (-2,1,0) \) with direction \( (3,0,4) \) (magnitude 5).
2. Vector from that point to A: \( (3,-2,2) \) cross with direction: \( (-8,-6,6) \) (magnitude \( \sqrt{136} \)). Height \( h=\frac{\sqrt{136}}{5} \).
3. Area \( =\frac12\cdot5\cdot\frac{\sqrt{136}}{5}=\frac{\sqrt{136}}{2}=\sqrt{34} \).

Teacher's Note:
1. Area = half base times perpendicular distance of A from the line.
2. Perpendicular distance from a point to a line uses a cross product.

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

Explanation:
1. A point on the line: \( (2s+1,-s-1,s) \). The foot is \( Q=(2s+1+k,\,-s-1+k,\,s+k) \) (moving along the normal \( (1,1,1) \)).
2. Q on \( x+y+z=3 \): \( (2s+1)+(-s-1)+s+3k=2s+3k=3 \).
3. Q on \( x-y+z=3 \): \( (2s+1+k)-(-s-1+k)+(s+k)=4s+2+k=3 \), so \( k=1-4s \).
4. Substituting: \( 2s+3-12s=3 \) gives \( s=0,\ k=1 \), so \( Q=(2,0,1) \).

Teacher's Note:
1. Move from a general point of the line along the plane's normal.
2. Use both plane equations to find the two unknowns.

Answer: (d) 3

Explanation:
1. For the lines to be coplanar, \( \begin{vmatrix}-2&-5&2\\2&4&3\\2&6&\lambda\end{vmatrix}=0 \) gives \( \lambda=-7 \).
2. Normal \( (2,4,3)\times(2,6,-7)=(-46,20,4) \) is parallel to \( (23,-10,-2) \) the plane through \( (-1,3,-1) \) is \( 23x-10y-2z+51=0 \) (checked: \( -23-30+2+51=0 \)).
3. Distance between parallel planes \( =\frac{|51-48|}{\sqrt{633}}=\frac{3}{\sqrt{633}} \), so \( k=3 \).

Teacher's Note:
1. First fix lambda from the coplanarity condition.
2. The plane is parallel to the given plane, so only the constant term differs.

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

Explanation:
1. The vector from \( (0,2,3) \) to the point is \( (3,-3,8) \) and the direction is \( (2,3,4) \).
2. Squared distance \( =|\vec w|^2-\frac{(\vec w\cdot\vec d)^2}{|\vec d|^2}=82-\frac{(6-9+32)^2}{29}=82-29=53 \).
3. The distance is \( \sqrt{53} \).

Teacher's Note:
1. Perpendicular distance squared = |w|^2 minus the square of the component along the line.
2. Alternatively use |w x d| / |d|.

Answer: (a) \( x-2y+2z-3=0 \)

Explanation:
1. A parallel plane is \( x-2y+2z+d=0 \).
2. Distance from the origin \( \frac{|d|}{\sqrt{1+4+4}}=\frac{|d|}{3}=1 \), so \( d=\pm3 \).
3. Among the options only \( d=-3 \) qualifies.

Teacher's Note:
1. Parallel planes share the normal vector.
2. Distance from origin is |d| divided by the normal's length.

Answer: (a) \( \frac92 \)

Explanation:
1. Lines intersect if the vector joining points \( (1,-1,1) \) and \( (3,k,0) \) is coplanar with the directions: \( \begin{vmatrix}2&k+1&-1\\2&3&4\\1&2&1\end{vmatrix}=0 \).
2. Expanding: \( 2(3-8)-(k+1)(2-4)-1(4-3)=-10+2k+2-1=0 \), so \( k=\frac92 \).

Teacher's Note:
1. Two lines meet when the scalar triple product of the joining vector and the two directions is zero.
2. Expand the determinant along the first row.

Answer: (b) exactly two values

Explanation:
1. The joining vector from \( (2,3,4) \) to \( (1,4,5) \) is \( (-1,1,1) \).
2. Coplanarity: \( \begin{vmatrix}-1&1&1\\1&1&-k\\k&2&1\end{vmatrix}=0 \) which expands to \( k^2+3k=0 \).
3. So \( k=0 \) or \( k=-3 \): exactly two values.

Teacher's Note:
1. Use the scalar triple product condition.
2. The determinant gives a quadratic in k.

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

Explanation:
1. Direction \( (3,1,-5) \) and normal \( (2,-1,1) \): \( 6-1-5=0 \), so the line is parallel to the plane and its image has the same direction.
2. Image of the point \( (1,3,4) \): value of plane \( 2-3+4+3=6 \), so image \( =(1,3,4)-\frac{2\cdot6}{6}(2,-1,1)=(-3,5,2) \).
3. Image line: \( \frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5} \).

Teacher's Note:
1. Check whether the line is parallel to the plane.
2. Then reflect one point of the line.

Answer: (a) \( \frac\pi3 \)

Explanation:
1. Eliminate \( n=-(l+m) \): \( l^2=m^2+(l+m)^2 \) gives \( 2m(m+l)=0 \).
2. If \( m=0 \): ratios \( (1,0,-1) \). If \( m=-l \): ratios \( (1,-1,0) \).
3. Cosine of the angle \( =\frac{|1+0+0|}{\sqrt2\sqrt2}=\frac12 \), so the angle is \( \frac\pi3 \).

Teacher's Note:
1. Eliminate one variable to find the two direction ratio sets.
2. Use the dot product of the two sets.

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

Explanation:
1. Family: \( (2x-5y+z-3)+\mu(x+y+4z-5)=0 \) with normal \( (2+\mu,\,-5+\mu,\,1+4\mu) \).
2. Parallel to \( (1,3,6) \): \( -5+\mu=3(2+\mu) \) gives \( \mu=-\frac{11}{2} \).
3. Then the plane is \( -\frac72x-\frac{21}{2}y-21z+\frac{49}{2}=0 \), i.e. \( x+3y+6z=7 \).

Teacher's Note:
1. Use the pencil of planes through the line of intersection.
2. Make the normal proportional to (1, 3, 6).

Answer: (a) 2

Explanation:
1. The point \( (3,-2,-4) \) is on the plane: \( 3l-2m+4=9 \), i.e. \( 3l-2m=5 \).
2. The direction is perpendicular to the normal: \( 2l-m-3=0 \), i.e. \( m=2l-3 \).
3. Then \( 3l-4l+6=5 \) gives \( l=1,\ m=-1 \), so \( l^2+m^2=2 \).

Teacher's Note:
1. A line lies in a plane if one point lies on it and the direction is perpendicular to the normal.
2. Two equations in l and m.

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

Explanation:
1. Direction cosines satisfy \( \cos^2 45^\circ+\cos^2120^\circ+\cos^2\theta=1 \).
2. \( \frac12+\frac14+\cos^2\theta=1 \), so \( \cos^2\theta=\frac14 \).
3. For an acute angle, \( \cos\theta=\frac12 \) and \( \theta=60^\circ \).

Teacher's Note:
1. Use l^2 + m^2 + n^2 = 1.
2. Choose the acute angle.

Answer: (b) \( x-4y+7z=0 \)

Explanation:
1. Image: \( (3,1,7) \) gives \( 3-1+7-3=6 \), so \( P=(3,1,7)-\frac{2\cdot6}{3}(1,-1,1)=(-1,5,3) \).
2. The line passes through the origin with direction \( (1,2,1) \); the plane has normal \( \vec{OP}\times(1,2,1)=(-1,5,3)\times(1,2,1)=(-1,4,-7) \).
3. Plane through the origin: \( -x+4y-7z=0 \), i.e. \( x-4y+7z=0 \).

Teacher's Note:
1. Reflect the point using the formula for the image in a plane.
2. The plane contains the origin, so it has no constant term.

Answer: (b) \( 14x+2y+15z=31 \)

Explanation:
1. The normal is \( (2,1,-2)\times(3,-6,-2)=(-2-12,\,-6+4,\,-12-3)=(-14,-2,-15) \), i.e. \( (14,2,15) \).
2. Plane: \( 14x+2y+15z=14+2+15=31 \).

Teacher's Note:
1. A plane perpendicular to two planes has normal along the cross product of their normals.
2. Use the point to find the constant.

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

Explanation:
1. Direction of the line of intersection: \( (2,1,-1)\times(1,2,1)=(3,-3,3)\parallel(1,-1,1) \), so (i) is false.
2. (ii) The given line has direction \( (3,-3,3) \), parallel (not perpendicular) to it: false.
3. (iii) \( \cos\theta=\frac{2+2-1}{\sqrt6\sqrt6}=\frac12 \): true.
4. (iv) The plane is \( x-y+z=0 \) (through \( (4,2,-2) \): \( 4-2-2=0 \)); distance of \( (2,1,1) \) is \( \frac{2}{\sqrt3} \): true.

Teacher's Note:
1. Find the line direction by a cross product of normals.
2. Test each statement separately.

Answer: (c) \( \vec r=\frac29(4\hat i+\hat j+\hat k)+t(2\hat i+2\hat j-\hat k) \)

Explanation:
1. Direction of \( L_3 \): \( (-1,2,2)\times(2,-1,2)=(6,6,-3)\parallel(2,2,-1) \).
2. The common perpendicular joins \( (\tfrac89,\tfrac29,\tfrac29) \) on \( L_1 \) (at \( \lambda=\tfrac19 \)) to \( (\tfrac49,-\tfrac29,\tfrac49) \) on \( L_2 \) (at \( \mu=\tfrac29 \)).
3. The point \( \frac29(4,1,1)=(\tfrac89,\tfrac29,\tfrac29) \) lies on it, so \( \vec r=\frac29(4\hat i+\hat j+\hat k)+t(2\hat i+2\hat j-\hat k) \) is correct. The line through the origin is not (it does not meet \( L_1 \)).

Teacher's Note:
1. The common perpendicular has direction equal to the cross product.
2. Find the feet by solving for the shortest-distance points.

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

Explanation:
1. Vertices (intersections of triples of planes): \( O(0,0,0) \), \( V_1(\sqrt3a,\sqrt3a,-\sqrt3a) \), \( V_2(\sqrt3a,-\sqrt3a,\sqrt3a) \), \( V_3(-\sqrt3a,\sqrt3a,\sqrt3a) \).
2. Take opposite edges \( OV_1 \) (direction \( (1,1,-1) \)) and \( V_2V_3 \) (direction \( (-1,1,0) \)). Their cross product is \( (1,1,2) \) with magnitude \( \sqrt6 \).
3. Distance \( =\frac{|\vec{OV_2}\cdot(1,1,2)|}{\sqrt6}=\frac{2\sqrt3\,a}{\sqrt6}=\sqrt2\,a \); by symmetry the other pairs give the same value.

Teacher's Note:
1. Find the four vertices first.
2. Use the shortest-distance formula for skew lines.

Answer: (d) \( \frac x\alpha+\frac y\beta+\frac z\gamma=3 \)

Explanation:
1. Let the intercepts be \( p,q,r \): \( A(p,0,0),B(0,q,0),C(0,0,r) \).
2. Centroid \( =\left(\frac p3,\frac q3,\frac r3\right)=(\alpha,\beta,\gamma) \), so \( p=3\alpha,\ q=3\beta,\ r=3\gamma \).
3. Plane \( \frac xp+\frac yq+\frac zr=1 \) becomes \( \frac{x}{3\alpha}+\frac{y}{3\beta}+\frac{z}{3\gamma}=1 \), i.e. \( \frac x\alpha+\frac y\beta+\frac z\gamma=3 \).

Teacher's Note:
1. Use the intercept form of the plane.
2. Express intercepts through the centroid.

Answer: (c) -2

Explanation:
1. Two lines, each given as an intersection of two planes, are coplanar when the determinant of the four planes (coefficients and constants) vanishes: \( \begin{vmatrix}2&-1&3&4\\a&1&-1&2\\1&-3&1&0\\1&2&1&1\end{vmatrix}=0 \).
2. Expanding gives \( 12(a+2)=0 \), so \( a=-2 \).

Teacher's Note:
1. Use the four-plane determinant condition.
2. The determinant factorises as 12(a + 2).

Answer: (a) \( \frac{35\sqrt2}{12} \)

Explanation:
1. The planes are parallel, at perpendicular distance \( \frac{|-12-2|}{\sqrt{4+9+36}}=\frac{14}{7}=2 \).
2. The cosine of the angle between the line and the plane normal is \( \frac{|6-12+30|}{7\cdot5\sqrt2}=\frac{24}{35\sqrt2} \).
3. \( PQ=\frac{2}{\cos\phi}=\frac{2\cdot35\sqrt2}{24}=\frac{35\sqrt2}{12} \).

Teacher's Note:
1. PQ times the cosine of the angle with the normal equals the perpendicular gap.
2. Compute the gap between the planes first.

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

Explanation:
1. Rotation by \( 90^\circ \) about the z-axis sends the x-axis to the y-axis and keeps z fixed, so the direction \( (1,0,1) \) becomes \( (0,1,1) \).
2. \( \cos\theta=\frac{0+0+1}{\sqrt2\sqrt2}=\frac12 \), so \( \theta=60^\circ \).

Teacher's Note:
1. Track where the x-axis goes under the rotation.
2. Use the dot product of the two directions.

Answer: (a) \( \frac23 \)

Explanation:
1. Vertices: \( O(0,0,0) \) and the points where the plane \( x+y+z=1 \) meets pairs of the three planes: \( (1,1,-1),\ (1,-1,1),\ (-1,1,1) \).
2. Volume \( =\frac16\left|\begin{vmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{vmatrix}\right|=\frac16\cdot4=\frac23 \).

Teacher's Note:
1. Find the four vertices.
2. Volume = one sixth of the triple product of three edges from one vertex.

Answer: (b) \( \frac74 \)

Explanation:
1. Direction of the line of intersection: \( (2,1,0)\times(1,-1,1)=(1,-2,-3) \).
2. Direction of \( L \) is perpendicular to this and to the plane's normal \( (4,1,1) \): \( (1,-2,-3)\times(4,1,1)=(1,-13,9) \).
3. The point \( (3,\alpha,\beta) \) gives \( (1,\alpha+1,\beta+1)\parallel(1,-13,9) \): \( \alpha=-14,\ \beta=8 \).
4. Hence \( \left|\frac\alpha\beta\right|=\frac{14}{8}=\frac74 \).

Teacher's Note:
1. Find the direction of L from two perpendicularity conditions.
2. Match the vector from (2,-1,-1) to (3, alpha, beta) with the direction.

Answer: (d) 32

Explanation:
1. The line lies in the plane (direction check \( 6-12+6=0 \) and point check \( 3+8-18+7=0 \)), so the rotating plane always contains the line.
2. The new plane contains the line and the origin, so its normal is \( (1,2,-3)\times(2,-3,1)=(-7,-7,-7) \), giving the plane \( x+y+z=0 \), i.e. \( 4x+4y+4z=0 \).
3. So \( \alpha=\beta=4 \) and \( \alpha^2+\beta^2=32 \).

Teacher's Note:
1. Check that the line lies in the original plane.
2. The new plane is determined by the line and the origin.