JEE Mathematics Three Dimensional Geometry MCQs Set 23

Answer: (a) \( \frac67,\frac27,\frac37 \)

Explanation:
1. In a parallelogram \( \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=(7,5,5)+(5,-1,1)=(12,4,6) \).
2. Magnitude \( \sqrt{144+16+36}=14 \).
3. Direction cosines \( \frac{12}{14},\frac{4}{14},\frac{6}{14}=\frac67,\frac27,\frac37 \).

Teacher's Note:
1. The diagonal from \( A \) is the sum of the two adjacent sides from \( A \).
2. Direction cosines are the components divided by the magnitude.

Answer: (b) \( (2,-1,-1) \) and \( (1,1,-2) \)

Explanation:
1. Put \( n=-l-m \) in the second relation: \( 2m(-l-m)+2ml+l(l+m)=0 \) gives \( l^2+lm-2m^2=0 \), i.e. \( (l-m)(l+2m)=0 \).
2. If \( l=m \) then \( n=-2l \): ratios \( (1,1,-2) \). If \( l=-2m \) then \( n=m \): ratios \( (2,-1,-1) \).
3. Both satisfy the two given relations.

Teacher's Note:
1. Eliminate one variable using the linear relation and factorise the resulting quadratic.
2. Each factor gives one line.

Answer: (b) \( 2,\ 8 \)

Explanation:
1. Points: \( (1,2,0) \) and \( (-7,0,-\lambda) \); join \( (-8,-2,-\lambda) \).
2. Coplanarity: \( \begin{vmatrix}4&1&1\\\lambda&\lambda-6&2\\-8&-2&-\lambda\end{vmatrix}=0 \).
3. Expanding gives \( \lambda^2-10\lambda+16=0 \), so \( \lambda=2,8 \).

Teacher's Note:
1. Two lines are coplanar iff the join of two points and the two directions have zero triple product.
2. Expand carefully and factorise the quadratic.

Answer: (b) \( \frac{3}{\sqrt{114}}(-7\hat i+8\hat j-\hat k) \)

Explanation:
1. Write \( \vec v=\alpha(1,1,1)+\beta(2,-3,0) \) with \( \vec v\cdot(1,1,1)=3\alpha-\beta=0 \), so \( \beta=3\alpha \).
2. Then \( \vec v\propto(7,-8,1) \), of magnitude \( \sqrt{114} \).
3. A vector of length 3 is \( \pm\frac{3}{\sqrt{114}}(7,-8,1) \); the option \( \frac{3}{\sqrt{114}}(-7,8,-1) \) is the negative one listed.

Teacher's Note:
1. A vector in a plane is a combination of the two spanning vectors.
2. Impose perpendicularity, then scale to length 3.

Answer: (b) \( \left(\frac12,\frac52,0\right) \)

Explanation:
1. Points on the line: \( (3-5t,\ 2+t,\ 5-10t) \).
2. xy-plane means \( z=0 \): \( t=\frac12 \).
3. Point \( \left(\frac12,\frac52,0\right) \).

Teacher's Note:
1. Parametrise the line and set the coordinate that vanishes on the plane to zero.
2. Substitute the parameter back into \( x \) and \( y \).

Answer: (b) \( (1,3,5) \)

Explanation:
1. A general point is \( F(t,\ 1+2t,\ 2+3t) \). Then \( \overrightarrow{PF}=(t-1,\ 2t-5,\ 3t-1) \).
2. Perpendicular to the direction \( (1,2,3) \): \( (t-1)+2(2t-5)+3(3t-1)=14t-14=0 \), so \( t=1 \).
3. Foot \( (1,3,5) \).

Teacher's Note:
1. Write the foot as a point on the line with a parameter.
2. The join from the external point to the foot is perpendicular to the line direction.

Answer: (b) \( 3\sqrt{30} \)

Explanation:
1. Directions \( (3,-1,1) \) and \( (-3,2,4) \); cross product \( (-6,-15,3) \).
2. Join of given points \( (-6,-15,3) \), which is exactly parallel to the cross product, so the given points themselves are the feet: \( A(3,8,3),\ B(-3,-7,6) \).
3. Distance \( =\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30} \).

Teacher's Note:
1. The common perpendicular is along \( d1 x d2 \).
2. If the join of the given points is parallel to \( d1 x d2 \) it is already the common perpendicular.

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

Explanation:
1. Normal \( (1,1,-1)\times(2,-1,0)=(-1,-2,-3) \), so the plane is \( x+2y+3z=d \).
2. Through \( (0,1,0) \): \( d=2 \). Plane \( x+2y+3z-2=0 \).

Teacher's Note:
1. A plane containing two direction vectors has their cross product as normal.
2. Substitute the given point to find the constant.

Answer: (d) \( x+y+z=6 \)

Explanation:
1. Equal intercepts: \( x+y+z=a \).
2. Through \( (1,2,3) \): \( a=6 \).

Teacher's Note:
1. Equal intercepts mean the coefficients of \( x, y, z \) are equal.
2. Substitute the point.

Answer: (c) \( 3:2,\ \left(0,\frac{13}{5},2\right) \)

Explanation:
1. Points on the line: \( (3-5t,\ 5-4t,\ -7+15t) \).
2. yz-plane \( x=0 \): \( t=\frac35 \), so the ratio is \( 3:2 \).
3. Point \( \left(0,\frac{13}{5},2\right) \).

Teacher's Note:
1. The yz-plane is \( x = 0 \).
2. If \( t \) is the fraction of the way, the ratio is \( t \) : \( (1 - t) \).

Answer: (b) \( x^2+y^2+z^2=9 \)

Explanation:
1. The three normals \( (1,1,1),(1,0,-1),(1,-2,1) \) are mutually perpendicular.
2. Sum of squared distances \( =\frac{(x+y+z)^2}{3}+\frac{(x-z)^2}{2}+\frac{(x-2y+z)^2}{6} \).
3. This simplifies to \( x^2+y^2+z^2 \), so the locus is \( x^2+y^2+z^2=9 \).

Teacher's Note:
1. Mutually perpendicular planes act like coordinate planes (rotated).
2. The distance from a plane is |ax+by+cz+d| divided by the length of the normal.

Answer: (a) \( \frac{x-4}{9}=\frac{y+1}{-1}=\frac{z-7}{-3} \)

Explanation:
1. Direction \( (9,-1,-3) \) with normal \( (3,-3,10) \): \( 27+3-30=0 \), so the line is parallel to the plane and keeps its direction.
2. Reflect the point \( (1,2,-3) \): \( 3-6-30-26=-59 \) and \( |\vec n|^2=118 \), so the image is \( (1,2,-3)+(3,-3,10)=(4,-1,7) \).
3. The image line is \( \frac{x-4}{9}=\frac{y+1}{-1}=\frac{z-7}{-3} \).

Teacher's Note:
1. Check whether the line is parallel to the plane, then reflect one point.
2. The image of a point \( P \) is \( P \) minus 2 (value / |n|^2) n.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. Vectors in the plane: \( (0,-2,0) \) and \( (-8,-4,-6) \). Normal \( (12,0,-16)\parallel(3,0,-4) \).
2. The xz-plane has normal \( (0,1,0) \), and \( (3,0,-4)\cdot(0,1,0)=0 \).
3. The normals are perpendicular, so the angle between the planes is \( \frac\pi2 \).

Teacher's Note:
1. The angle between planes equals the angle between their normals.
2. A zero dot product of normals means perpendicular planes.

Answer: (a) \( 7 \)

Explanation:
1. Points: \( (2+3t,\ 3+6t,\ 4+2t) \). On the plane: \( 3(2+3t)+2(3+6t)+2(4+2t)+5=25t+25=0 \), so \( t=-1 \).
2. Distance \( =|t|\sqrt{3^2+6^2+2^2}=7 \).

Teacher's Note:
1. Distance along a direction is \( |t| \) times the length of the direction vector.
2. Find \( t \) by substituting the parametric point into the plane.

Answer: (c) \( x-y+3z-2=0 \) and \( \sqrt{11} \)

Explanation:
1. Normal \( (2-1,\ 3-4,\ 5-2)=(1,-1,3) \). Plane through \( (1,2,1) \): \( x-y+3z=2 \).
2. Distance of \( (4,0,3) \): \( \frac{|4-0+9-2|}{\sqrt{11}}=\frac{11}{\sqrt{11}}=\sqrt{11} \).

Teacher's Note:
1. A plane perpendicular to a line has the line's direction as normal.
2. Distance = |value of the plane expression| / |normal|.

Answer: (c) \( \left(\frac{76}{35},-\frac{108}{35},-\frac{170}{35}\right) \)

Explanation:
1. The line is \( \vec r=(2,-3,-5)+t(6,-3,5) \).
2. Put into the plane \( 6x-3y+5z=-2 \): \( 12+36t+9+9t-25+25t=-2 \), so \( 70t=2,\ t=\frac1{35} \).
3. Point \( \left(2+\frac6{35},-3-\frac3{35},-5+\frac5{35}\right)=\left(\frac{76}{35},-\frac{108}{35},-\frac{170}{35}\right) \).

Teacher's Note:
1. A line perpendicular to a plane runs along the plane's normal.
2. Substitute the parametric point into the plane equation.

Answer: (b) \( \frac53 \)

Explanation:
1. Rewrite the second plane with the same normal: \( 2x-3y+6z+\frac{20}{3}=0 \) and the first as \( 2x-3y+6z-5=0 \).
2. Distance \( =\frac{|\frac{20}{3}+5|}{\sqrt{4+9+36}}=\frac{35/3}{7}=\frac53 \).

Teacher's Note:
1. Make the normals identical before applying the parallel-plane distance formula.
2. Distance = \( |d1 - d2| \) / |normal|.

Answer: (c) \( -1 \)

Explanation:
1. The first two lines pass through \( (1,2,3) \).
2. The third passes through \( (\lambda^2,2,3) \), which must equal \( (1,2,3) \) for concurrency: \( \lambda^2=1 \).
3. If \( \lambda=1 \) the third line has direction \( (3,2,1) \), the same as the first, so the lines would not be distinct. Hence \( \lambda=-1 \).

Teacher's Note:
1. Concurrent lines share a common point.
2. Reject any value that makes two of the lines identical.

Answer: (a) \( \frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2} \) and \( \frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6} \)

Explanation:
1. Each equal side makes \( 45^\circ \) with the hypotenuse direction \( (5,3,8) \) (magnitude \( 7\sqrt2 \)).
2. For \( (3,6,2) \): \( \cos\theta=\frac{15+18+16}{7\cdot7\sqrt2}=\frac1{\sqrt2} \). For \( (2,-3,6) \): \( \frac{10-9+48}{7\cdot7\sqrt2}=\frac1{\sqrt2} \).
3. Both lines pass through the vertex \( (7,2,4) \), so both are sides.

Teacher's Note:
1. In an isosceles right triangle each leg makes 45 degrees with the hypotenuse.
2. Test each candidate direction with the dot product.

Answer: (b) \( x+5y-6z+19=0 \)

Explanation:
1. Normal \( (3-2,\ 4+1,\ -1-5)=(1,5,-6) \).
2. Plane \( x+5y-6z=2-15-6=-19 \), i.e. \( x+5y-6z+19=0 \).

Teacher's Note:
1. The line's direction is the plane's normal.
2. Evaluate the left side at the given point to get the constant.

Answer: (b) \( 13 \)

Explanation:
1. Point on the line: \( (2+3t,\ -1+4t,\ 2+12t) \). On the plane: \( 5+11t=5 \), so \( t=0 \).
2. Intersection \( (2,-1,2) \); distance to \( (-1,-5,-10) \) is \( \sqrt{9+16+144}=13 \).

Teacher's Note:
1. Find the intersection point first, then use the distance formula.
2. Here the given point on the line already lies on the plane.

Answer: (b) \( 1 \)

Explanation:
1. Point \( (1+2t,\ -2+3t,\ 3-6t) \) on the plane: \( 1+2t+2-3t+3-6t=6-7t=5 \), so \( t=\frac17 \).
2. Distance \( =|t|\sqrt{4+9+36}=\frac17\cdot7=1 \).

Teacher's Note:
1. Distance along a direction = \( |t| x \) length of the direction vector.
2. Solve the plane equation for \( t \).