JEE Mathematics Three Dimensional Geometry MCQs Set 17

Answer: (d) \( \frac{|\vec b\cdot\vec n|}{|\vec b||\vec n|} \)

Explanation:
1. The angle between a line and a plane is the complement of the angle between the line and the normal.
2. So \( \sin\theta=\cos(\text{angle between }\vec b\text{ and }\vec n)=\frac{|\vec b\cdot\vec n|}{|\vec b||\vec n|} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) \( \hat i+2\hat j+\hat k \)

Explanation:
1. First line: \( (5-4\lambda,1+\lambda,2-\lambda) \); second: \( (-1+2\mu,7-5\mu,8-7\mu) \).
2. Taking \( \lambda=\mu=1 \) gives (1, 2, 1) on both lines.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( 9\hat i+5\hat j-2\hat k \)

Explanation:
1. \( (2+\lambda,6+3\lambda,3+5\lambda)\cdot(1,3,5)=35+35\lambda=0 \), so \( \lambda=-1 \) and the foot is (8, 2, 0).
2. Image \( =2(8,2,0)-(7,-1,2)=(9,5,-2) \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( 10x-12y-9z+60=0 \)

Explanation:
1. The vector from (-3, 1, 2) to (0, 2, 4) is (3, 1, 2); the normal is \( (3,4,-2)\times(3,1,2)=(10,-12,-9) \).
2. The plane is \( 10x-12y-9z=-24-36 \), i.e. \( 10x-12y-9z+60=0 \).
3. Check with (-3, 1, 2): \( -30-12-18+60=0 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( 13x-23y+32z+59=0 \)

Explanation:
1. \( \frac{2x-3y+6z+8}7=\pm\frac{x-2y+2z+5}3 \).
2. The plus sign gives \( 3(2x-3y+6z+8)+7(x-2y+2z+5)=13x-23y+32z+59 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) intersect at (4, 0, -1)

Explanation:
1. A point of the first line is \( (1+3t,1-t,-1) \); for t = 1 it is (4, 0, -1).
2. The second line passes through (4, 0, -1) with parameter 0.
3. So they meet at (4, 0, -1).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) 3

Explanation:
1. The y-coordinate of the dividing point is \( \frac{-\lambda+3}{\lambda+1} \).
2. It is zero on the XZ-plane, so \( \lambda=3 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( \frac{11}{\sqrt{107}} \)

Explanation:
1. A point of the line is (1, 0, -1).
2. The plane value is \( 5+0-9-7=-11 \) and \( |n|=\sqrt{107} \).
3. Distance \( \frac{11}{\sqrt{107}} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) meet in a unique point

Explanation:
1. The directions (3, -1, 0) and (2, 0, 3) are not proportional, so the lines are not parallel.
2. The difference of the points is (3, -1, 0) and \( (3,-1,0)\cdot[(3,-1,0)\times(2,0,3)]=0 \), so the lines are coplanar.
3. Non-parallel coplanar lines meet in exactly one point.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) 2

Explanation:
1. \( r^2=4\sin^2t+4\cos^2t+9t^2=4+9t^2 \).
2. The minimum is at t = 0, giving \( r^2=4 \), i.e. 2.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) A-R, B-P, C-S, D-Q

Explanation:
1. \( \text{origin: }\frac{10}7 \); \( (3,6,2):\ \frac{|9-36+4+10|}7=\frac{13}7 \).
2. \( (2,-3,6):\ \frac{6+18+12+10}7=\frac{46}7 \); \( (-6,2,3):\ \frac{|-18-12+6+10|}7=2 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) 5

Explanation:
1. \( \lambda^2=4^2+5^2=41 \) (the distance from the y-axis uses the x and z coordinates).
2. Hence \( \frac{5\cdot41}{41}=5 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

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

Explanation:
1. The normal is (0, 1, 1); its dot product with the x-direction (1, 0, 0) is zero, so the plane is parallel to the x-axis and A is true.
2. The normal is perpendicular to the x-axis, not parallel, so R is false.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) \( x^2+y^2+z^2+28x-12y+10z-247=0 \)

Explanation:
1. \( 9PA^2=4PB^2 \): \( 9[(x+2)^2+(y-2)^2+(z-3)^2]=4[(x-13)^2+(y+3)^2+(z-13)^2] \).
2. Expanding: \( 5(x^2+y^2+z^2)+140x-60y+50z-1235=0 \).
3. Divide by 5.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) \( x-3y-5z+20=0 \)

Explanation:
1. The normal is \( (2,-1,1)\times(1,2,-1)=(-1,3,5) \).
2. The plane is \( -x+3y+5z=1+9+10=20 \), i.e. \( x-3y-5z+20=0 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) 27

Explanation:
1. Plane \( \frac xa+\frac yb+\frac zc=1 \) through (1, 2, 3): \( \frac1a+\frac2b+\frac3c=1 \).
2. By AM-GM \( 1\ge3\sqrt[3]{\frac{6}{abc}} \), so \( abc\ge162 \).
3. Minimum volume \( \frac{162}6=27 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( \left(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6}\right) \)

Explanation:
1. Put \( n=-l-m \): \( 2m(-l-m)+2ml+l(l+m)=l^2+lm-2m^2=(l-m)(l+2m)=0 \).
2. So l = m, n = -2m, giving \( \left(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6}\right) \) (the other solution is l = -2m).
3. The other options do not satisfy l + m + n = 0.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) 7

Explanation:
1. The point \( (4,2,k) \) must satisfy the plane: \( 8-8+k=7 \), so k = 7.
2. The direction check \( 2-4+2=0 \) holds, so the line is parallel to the plane and lies in it.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) 4

Explanation:
1. The circumcentre lies in the plane of the triangle, which is \( x+y+z=0 \) (the normal is \( (-6,6,0)\times(-6,0,6)\propto(1,1,1) \) and the plane passes through (3, 2, -5)).
2. So \( -1+\lambda-3=0 \) and \( \lambda=4 \).
3. Check: the distance squared from (-1, 4, -3) to each vertex is 24.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) (-3, -2, 13)

Explanation:
1. The lines meet at \( (5,-2,-1) \) (parameters s = 2, t = 1).
2. The normal is \( (7,1,2)\times(3,5,-6)=(-16,48,32)\propto(-1,3,2) \), i.e. (1, -3, -2) after sign change.
3. The plane is \( x-3y-2z=5+6+2=13 \), so (b, c, d) = (-3, -2, 13).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (d) \( \tan^{-1}\frac52 \)

Explanation:
1. Normals: \( \vec{AD}\times\vec{AC}=(6,3,-6) \) and \( \vec{AB}\times\vec{AC}=(-2,4,-3) \).
2. \( \cos\theta=\frac{|-12+12+18|}{9\sqrt{29}}=\frac{2}{\sqrt{29}} \).
3. So \( \tan\theta=\frac52 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (c) \( x^2+y^2+z^2-xy-yz-zx-2=0 \)

Explanation:
1. \( |\vec\omega\times\vec r|=10\sqrt3\sqrt{r^2-\frac{(x+y+z)^2}3}=20 \).
2. So \( 3(x^2+y^2+z^2)-(x+y+z)^2=4 \).
3. This is \( 2(x^2+y^2+z^2-xy-yz-zx)=4 \), i.e. the first option.

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (b) 7

Explanation:
1. \( AB=(1,-3,-2),AC=(0,0,-7) \); \( AB\times AC=(21,7,0) \) with magnitude \( 7\sqrt{10} \).
2. \( \lambda=\frac{7\sqrt{10}}2 \), so \( \frac{2\lambda}{\sqrt{10}}=7 \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.

Answer: (a) \( \frac{17}{5\sqrt3} \)

Explanation:
1. Points: (-1, -2, -1) and (2, -2, 3), difference (3, 0, 4).
2. \( d=\frac{|(3,0,4)\cdot(-1,-7,5)|}{5\sqrt3}=\frac{17}{5\sqrt3} \).

Teacher's Note:
1. Write down the standard form of the line or plane first and read off direction ratios or normals.
2. Check the final result against a given point or direction before choosing an option.