JEE Mathematics Three Dimensional Geometry MCQs Set 06

Answer: (d) (−7, 11, −4)

Explanation:
1. A general point is \( (1+2t,\ -1-3t,\ t) \); its distance from (1, −1, 0) is \( \sqrt{4t^2+9t^2+t^2}=\sqrt{14}\,|t| \).
2. Setting this equal to \( 4\sqrt{14} \): \( t=\pm4 \), giving (9, −13, 4) and (−7, 11, −4).
3. Distances from the origin: \( \sqrt{81+169+16} \) for the first and \( \sqrt{49+121+16} \) for the second. The nearer one is (−7, 11, −4).

Teacher's Note:
1. Use the parametric form of the line.
2. Compare the two points' distances from the origin.

Answer: (b) \( \sqrt{\frac{14}3} \)

Explanation:
1. Equal angles with the axes: direction ratios \( (1,1,1) \), with unit vector \( \frac{(1,1,1)}{\sqrt3} \).
2. Vector \( \vec{PA}=(1,-2,-1) \). The cross product with \( (1,1,1) \) is \( (-2+1,\ -1-1,\ 1+2)=(-1,-2,3) \), of magnitude \( \sqrt{14} \).
3. Distance \( =\frac{\sqrt{14}}{\sqrt3}=\sqrt{\frac{14}3} \).

Teacher's Note:
1. Direction ratios of a line equally inclined to the axes are equal.
2. Distance = |PA × d|/|d|.

Answer: (b) (3, −2, 1)

Explanation:
1. A point on the line is \( (1+2s,\ 1-3s,\ 2-s) \).
2. Substituting in the plane: \( 3+6s+2-6s+2-s=6 \), so \( 7-s=6 \) and s = 1.
3. The point is (3, −2, 1).

Teacher's Note:
1. Parametrise the line.
2. Substitute into the equation of the plane.

Answer: (c) \( \left(\frac a2,\frac b2,\frac c2\right) \)

Explanation:
1. Let the point be (x, y, z). Equal distances from (0, 0, 0) and (a, 0, 0): \( x^2+y^2+z^2=(x-a)^2+y^2+z^2 \), so \( x=\frac a2 \).
2. Similarly \( y=\frac b2 \) and \( z=\frac c2 \).
3. The point is \( \left(\frac a2,\frac b2,\frac c2\right) \).

Teacher's Note:
1. Equate distances to the origin and to each of the other three points.
2. Each equation fixes one coordinate.

Answer: (a) \( -\frac92,9,9 \)

Explanation:
1. The plane passes through the mid-point \( (-1,3,4) \) with normal along \( (-4,2,2)\propto(-2,1,1) \).
2. Equation: \( -2(x+1)+(y-3)+(z-4)=0 \), i.e. \( -2x+y+z=9 \).
3. Intercepts: \( x=-\frac92,\ y=9,\ z=9 \).

Teacher's Note:
1. The plane passes through the mid-point and is perpendicular to the joining line.
2. Divide by the constant term to get the intercept form.

Answer: (c) \( \frac{x-1}2=\frac y3=\frac z5 \)

Explanation:
1. Direction ratios: the cross product of the normals \( (1,1,-1)\times(2,-3,1)=(1-3,\ -2-1,\ -3-2)=(-2,-3,-5) \), i.e. (2, 3, 5).
2. Put z = 0: \( x+y=1,\ 2x-3y=2 \), giving (1, 0, 0) on the line.
3. So the line is \( \frac{x-1}2=\frac y3=\frac z5 \).

Teacher's Note:
1. Direction = cross product of the two normals.
2. Find one point on the line by setting a variable to zero.

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

Explanation:
1. A line perpendicular to a plane is parallel to its normal \( (1,-2,-3) \).
2. Through (1, 1, −1): \( \frac{x-1}1=\frac{y-1}{-2}=\frac{z+1}{-3} \).

Teacher's Note:
1. The direction ratios are the coefficients of the plane.
2. Write the symmetric form.

Answer: (b) 17

Explanation:
1. The projections on the axes are the differences of coordinates, so the length is \( \sqrt{9^2+12^2+8^2}=\sqrt{81+144+64}=\sqrt{289}=17 \).

Teacher's Note:
1. Length² = sum of squares of the projections on the axes.
2. 289 = 17².

Answer: (b) have a line in common

Explanation:
1. Three times the second plane minus the third: \( (6x+9y+15z)-(6x+5y+9z)=15-10 \), i.e. \( 4y+6z=5 \), which is the first plane.
2. So the first plane's equation follows from the other two; the planes are dependent and consistent: they share a common line.

Teacher's Note:
1. Look for a linear relation between the three equations.
2. Check whether the relation also holds for the right-hand sides.

Answer: (b) \( \left(\frac12,1,-\frac32\right) \)

Explanation:
1. A point on the line: \( Q=(-1+3s,\ 2-2s,\ -1-s) \). Then \( \vec{PQ}=(-2+3s,\ 2-2s,\ -3-s) \).
2. Perpendicularity to the direction (3, −2, −1): \( 3(-2+3s)-2(2-2s)-(-3-s)=0 \), i.e. \( 14s-7=0 \), so \( s=\frac12 \).
3. The foot is \( \left(\frac12,1,-\frac32\right) \).

Teacher's Note:
1. Take a general point of the line.
2. PQ must be perpendicular to the direction of the line.

Answer: (a) 2x + y + 2 = 0

Explanation:
1. The projecting plane contains the line (point (−1, 0, 1), direction \( (-1,2,3) \)) and the normal \( (1,-2,1) \) of the given plane.
2. Its normal is \( (-1,2,3)\times(1,-2,1)=(2+6,\ 3+1,\ 2-2)=(8,4,0)\propto(2,1,0) \).
3. Through (−1, 0, 1): \( 2(x+1)+y=0 \), i.e. 2x + y + 2 = 0.

Teacher's Note:
1. The plane through the line perpendicular to the given plane gives the projection.
2. Normal = direction × normal of the plane.

Answer: (d) \( \sqrt{30} \)

Explanation:
1. Let Q be the point of the line \( (1+2t,\ -1+t,\ 3-3t) \) such that PQ is parallel to the plane: \( \vec{PQ}=(2t-1,\ t-2,\ 5-3t) \).
2. Perpendicular to the normal (1, 2, 1): \( (2t-1)+2(t-2)+(5-3t)=t=0 \).
3. So Q = (1, −1, 3) and \( PQ=\sqrt{1+4+25}=\sqrt{30} \).

Teacher's Note:
1. PQ parallel to the plane means PQ · normal = 0.
2. Find t, then the distance.

Answer: (b) \( 4\sqrt3 \)

Explanation:
1. Directions \( \vec d_1=(2,-7,5) \), \( \vec d_2=(2,1,-3) \): \( \vec d_1\times\vec d_2=(21-5,\ 10+6,\ 2+14)=(16,16,16) \).
2. Points (3, −15, 9) and (−1, 1, 9): the join is \( (-4,16,0) \).
3. Distance \( =\frac{|(-4,16,0)\cdot(16,16,16)|}{16\sqrt3}=\frac{192}{16\sqrt3}=4\sqrt3 \).

Teacher's Note:
1. Shortest distance = |(a₂ − a₁) · (d₁ × d₂)| / |d₁ × d₂|.
2. Simplify using (1, 1, 1) as the cross product direction.

Answer: (a) \( \frac{x-1}{-3}=\frac{y-2}5=\frac{z-3}4 \)

Explanation:
1. The direction is perpendicular to both normals: \( (1,-1,2)\times(3,1,1)=(-1-2,\ 6-1,\ 1+3)=(-3,5,4) \).
2. The line through (1, 2, 3): \( \frac{x-1}{-3}=\frac{y-2}5=\frac{z-3}4 \).

Teacher's Note:
1. A line parallel to two planes is perpendicular to both normals.
2. Take the cross product of the normals.

Answer: (b) (3, −2, 1)

Explanation:
1. A point on the line is \( (-3+3s,\ 2-2s,\ -1+s) \).
2. Substituting in the plane: \( 4(-3+3s)+5(2-2s)+3(-1+s)-5=0 \), i.e. \( 5s-10=0 \), so s = 2.
3. The point is (3, −2, 1).

Teacher's Note:
1. Parametrise the line and substitute into the plane.
2. Check the point against the plane.

Answer: (d) 60° or 120°

Explanation:
1. \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \): \( \frac12+\frac14+\cos^2\gamma=1 \), so \( \cos^2\gamma=\frac14 \).
2. Thus \( \cos\gamma=\pm\frac12 \) and the angle is 60° or 120°.

Teacher's Note:
1. Use l² + m² + n² = 1 with direction cosines.
2. Both signs of cos γ are possible.

Answer: (d) (a, b, c)

Explanation:
1. Write the plane as \( \frac x{3a}+\frac y{3b}+\frac z{3c}=1 \): A = (3a, 0, 0), B = (0, 3b, 0), C = (0, 0, 3c).
2. The centroid is \( \left(\frac{3a}3,\frac{3b}3,\frac{3c}3\right)=(a,b,c) \).

Teacher's Note:
1. Convert to intercept form first.
2. Centroid = average of the vertices.

Answer: (c) k = 0 or −3

Explanation:
1. Point difference \( (1,4,5)-(2,3,4)=(-1,1,1) \); directions \( (1,1,-k) \) and \( (k,2,1) \).
2. Coplanarity: \( \begin{vmatrix}-1&1&1\\1&1&-k\\k&2&1\end{vmatrix}=-1(1+2k)-1(1+k^2)+1(2-k)=-k^2-3k=0 \).
3. So k = 0 or k = −3.

Teacher's Note:
1. Coplanar lines: the scalar triple product of the join and the two directions vanishes.
2. Expand the determinant.

Answer: (c) \( \pm\sqrt5 \)

Explanation:
1. A point of the line is \( (2+3s,\ -1+2s,\ 1-s) \); it lies in the plane z = 0 when s = 1, giving (5, 1, 0).
2. This point must satisfy \( xy=c^2 \): \( 5\cdot1=c^2 \).
3. So \( c=\pm\sqrt5 \).

Teacher's Note:
1. Find where the line meets z = 0.
2. Substitute the point in xy = c².

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

Explanation:
1. Lines are coplanar if \( \begin{vmatrix}1&4&-5\\2&1&-3\\1&-3&2\end{vmatrix}=1(2-9)-4(4+3)-5(-6-1)=-7-28+35=0 \) (the rows are the join of the points and the two directions). So A is true.
2. Setting the two lines' general points equal, \( 1+2x_1=2+x_2 \) etc. leads exactly to the system in R, which is consistent: x₁ = y₁ = 1 satisfies all three equations.
3. So R is true, and its consistency is the condition for the lines to meet, hence to be coplanar.

Teacher's Note:
1. Check coplanarity with the triple product.
2. Equating the general points of the two lines gives a system of three equations.

Answer: (c) A is true, but R is false.

Explanation:
1. Take A(1, 0, −2) on the line and direction \( \vec d=(1,-1,2) \). With P(1, 2, −4): \( \vec{AP}=(0,2,-2) \) and \( \vec{AP}\times\vec d=(4-2,\ -2-0,\ 0-2)=(2,-2,-2) \).
2. Perpendicular distance \( =\frac{\sqrt{12}}{\sqrt6}=\sqrt2 \), not 1: R is false.
3. Since \( \sqrt2<2 \), the sphere of radius 2 about P cuts the line in two points: A is true.

Teacher's Note:
1. Compute the perpendicular distance with the cross product.
2. Two points exist at distance 2 when the perpendicular distance is less than 2.