JEE Mathematics Three Dimensional Geometry MCQs Set 13

Answer: (a) \( \frac{x+2}{3}=\frac{y-1}{0}=\frac{z-3}{1} \)

Explanation:
1. D.r.'s: (1 + 2, 1 - 1, 4 - 3) = (3, 0, 1).
2. Through (-2, 1, 3): \( \frac{x+2}{3}=\frac{y-1}{0}=\frac{z-3}{1} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

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

Explanation:
1. The direction must be perpendicular to the normal (2, 3, 1).
2. For (-1, 1, -1): -2 + 3 - 1 = 0. The line passes through (1, -2, 3).
3. The other options either have the wrong point or a direction with non-zero dot product.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (c) \( x\cos\alpha+y\sin\alpha=0 \)

Explanation:
1. A plane through the z-axis has the form ax + by = 0.
2. Perpendicular to the line means the normal is along the direction (cos alpha, sin alpha, 0).
3. So \( x\cos\alpha+y\sin\alpha=0 \).

Teacher's Note:
1. A plane perpendicular to a line has the line's direction as its normal.
2. The z-axis passing through it forces c = d = 0.

Answer: (c) \( \vec r=\left(\frac13\hat i-\frac13\hat j+\hat k\right)+\lambda(\hat i+2\hat j+3\hat k) \)

Explanation:
1. Divide: \( \frac{x-\frac13}{\frac16}=\frac{y+\frac13}{\frac13}=\frac{z-1}{\frac12} \).
2. D.r.'s (1, 2, 3) and a point (1/3, -1/3, 1).
3. So \( \vec r=\left(\frac13\hat i-\frac13\hat j+\hat k\right)+\lambda(\hat i+2\hat j+3\hat k) \).

Teacher's Note:
1. Divide each bracket by its coefficient of the variable.
2. The denominators then give the direction.

Answer: (c) \( \frac{x-3}{14}=\frac{y+1}{2}=\frac{z}{15} \)

Explanation:
1. Direction: (3, -6, -2) x (2, 1, -2) = (12 + 2, -4 + 6, 3 + 12) = (14, 2, 15).
2. Put z = 0: 3x - 6y = 15 and 2x + y = 5 give x = 3, y = -1.
3. The line is \( \frac{x-3}{14}=\frac{y+1}{2}=\frac z{15} \).

Teacher's Note:
1. Direction by cross product; a point by putting one variable equal to zero.
2. Check that the point satisfies both planes.

Answer: (b) 2

Explanation:
1. D.r.'s are (-3, 2k, 2) and (3k, 1, 7).
2. Perpendicular: \( -9k+2k+14=0 \).
3. So -7k = -14 and k = 2.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) 3x + 4y + 5z = 7

Explanation:
1. A line is perpendicular to a plane when its direction is the normal.
2. The direction is (3, 4, 5), so the plane is \( 3x+4y+5z=7 \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) \( \cos^{-1}\left(\frac{19}{21}\right) \)

Explanation:
1. \( \cos\theta=\frac{3+4+12}{3\cdot7}=\frac{19}{21} \).
2. So the angle is \( \cos^{-1}\frac{19}{21} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (a) \( \frac53 \)

Explanation:
1. \( \sin\theta=\frac{|2-2+2\sqrt\lambda|}{3\sqrt{5+\lambda}}=\frac13 \).
2. So \( 2\sqrt\lambda=\sqrt{5+\lambda} \), i.e. \( 4\lambda=5+\lambda \).
3. Hence \( \lambda=\frac53 \).

Teacher's Note:
1. The sine of the angle between a line and a plane uses direction . normal.
2. Square both sides carefully and check lambda is positive.

Answer: (a) \( \cos^{-1}\left(\frac23\right) \)

Explanation:
1. Dot product: 8 + 2 + 8 = 18.
2. Lengths: 9 and 3.
3. \( \cos\theta=\frac{18}{27}=\frac23 \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

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

Explanation:
1. Dot product: 7 - 10 + 3 = 0.
2. So the lines are perpendicular.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) 9x - 2y - 5z + 4 = 0

Explanation:
1. Normal = (2, -1, 4) x (1, 2, 1) = (-1 - 8, 4 - 2, 4 + 1) = (-9, 2, 5).
2. The plane is 9x - 2y - 5z = k through (1, -1, 3): k = 9 + 2 - 15 = -4.
3. So 9x - 2y - 5z + 4 = 0.

Teacher's Note:
1. Take the line's direction and the other plane's normal as two vectors in the new plane.
2. Their cross product is the normal.

Answer: (a) -2

Explanation:
1. Direction of the line: (3, -2, 1) x (4, -3, 4) = (-8 + 3, 4 - 12, -9 + 8) = (-5, -8, -1).
2. Parallel to the plane: \( -10+8-m=0 \).
3. So m = -2.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (a) (-6, 7)

Explanation:
1. Direction perpendicular to normal: \( 3-15-2\alpha=0 \), so alpha = -6.
2. The point (2, 1, -2) lies on the plane: \( 2+3+2\alpha+\beta=0 \), so \( 5-12+\beta=0 \).
3. Hence beta = 7.

Teacher's Note:
1. Two conditions are needed: perpendicularity of direction and normal, and a point of the line on the plane.
2. Do them in order.

Answer: (a) (3, 1, -2)

Explanation:
1. Cross product of the normals (1, -1, 1) and (1, -3, 0): (0 + 3, 1 - 0, -3 + 1) = (3, 1, -2).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

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

Explanation:
1. A point on the line through P with d.r.'s (1, 1, 1): (1 + t, -5 + t, 9 + t).
2. On the plane: \( (1+t)-(-5+t)+(9+t)=15+t=5 \), so t = -10.
3. Distance \( =|t|\sqrt3=10\sqrt3 \).

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (a) (14, 1, 5)

Explanation:
1. AB = (6, 2, 3), of length 7; the unit vector is (6, 2, 3)/7.
2. A point at distance 14 beyond B is A + 2(6, 2, 3) = (14, 1, 5).
3. The other points: A - 2(6, 2, 3) = (-10, -7, -7), which is not among the options.

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (a) 7

Explanation:
1. Let Q = (1 + 2t, 3 + 4t, 2 + 3t) on the line. PQ = (2t - 2, 4t - 5, 3t).
2. PQ is parallel to the plane: \( 3(2t-2)+2(4t-5)-2(3t)=8t-16=0 \), so t = 2.
3. Q = (5, 11, 8), and \( PQ=\sqrt{4+9+36}=7 \).

Teacher's Note:
1. Parallel to a plane means PQ is perpendicular to the normal.
2. Use that to find t.

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

Explanation:
1. A point on the line: (2 + 3t, -1 + 2t, 1 - t).
2. z = 0 gives t = 1, i.e. the point (5, 1, 0).
3. Then \( xy=5=c^2 \), so \( c=\pm\sqrt5 \).

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (c) (9, -6, 1)

Explanation:
1. A point on the line: (3 + 3t, 2 - 4t, -1 + t).
2. In 2x + 4y + 3z + 3 = 0: 14 - 7t = 0, so t = 2.
3. In x + 2y + 3z = 0: 4 - 2t = 0, so t = 2 as well.
4. The point is (9, -6, 1).

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (d) 0

Explanation:
1. Directions (3, 4, 5) and (2, 3, 4); cross product (1, -2, 1).
2. Join vector (1, 1, 1): dot product 1 - 2 + 1 = 0.
3. The scalar triple product is zero, so the lines are coplanar (they intersect) and the distance is 0.

Teacher's Note:
1. Use the standard formula and keep careful track of signs while subtracting the point vectors.
2. Verify the final value by an independent check where possible.

Answer: (a) \( \frac3{\sqrt2} \)

Explanation:
1. b1 x b2 = (2, 1, 2) x (1, -1, 1) = (3, 0, -3), with magnitude \( 3\sqrt2 \).
2. a2 - a1 = (-1, 3, 2), and its dot product with the cross product is -3 - 6 = -9.
3. Distance \( =\frac{9}{3\sqrt2}=\frac3{\sqrt2} \).

Teacher's Note:
1. Use the standard formula and keep careful track of signs while subtracting the point vectors.
2. Verify the final value by an independent check where possible.

Answer: (a) \( 2\sqrt{\frac{22}3} \)

Explanation:
1. a2 - a1 = (-5, 1, 2) and b = (1, -1, 2).
2. b x (a2 - a1) = (-4, -12, -4), with magnitude \( \sqrt{176}=4\sqrt{11} \).
3. Distance \( =\frac{4\sqrt{11}}{\sqrt6}=\sqrt{\frac{88}{3}}=2\sqrt{\frac{22}{3}} \).

Teacher's Note:
1. Use the standard formula and keep careful track of signs while subtracting the point vectors.
2. Verify the final value by an independent check where possible.

Answer: (b) (5, -8, -4)

Explanation:
1. Foot P = (1 + 2t, -1 - 3t, -10 + 8t). AP = (2t, -1 - 3t, -10 + 8t) is perpendicular to (2, -3, 8): \( 4t+3+9t-80+64t=77t-77=0 \), so t = 1.
2. P = (3, -4, -2).
3. The image is 2P - A = (5, -8, -4).

Teacher's Note:
1. The image is the point reflected through the foot of the perpendicular.
2. Find the foot first, then double it and subtract the point.

Answer: (c) 3r = a + b + c

Explanation:
1. The vector from (a, b, c) to (r, r, r) is perpendicular to (1, 1, 1): \( (r-a)+(r-b)+(r-c)=0 \).
2. So 3r = a + b + c.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (c) 7

Explanation:
1. Point on the line (-5, -3, 6); join vector (7, 7, -7).
2. Cross product with (1, 4, -9): (-35, 56, 21), squared magnitude 4802.
3. Distance \( =\sqrt{\frac{4802}{98}}=\sqrt{49}=7 \).

Teacher's Note:
1. Use the standard formula and keep careful track of signs while subtracting the point vectors.
2. Verify the final value by an independent check where possible.

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

Explanation:
1. Point on the line (0, 2, 3); join vector (3, -3, 8).
2. Cross product with (2, 3, 4): (-36, 4, 15), squared magnitude 1537.
3. \( \text{Distance}^2=\frac{1537}{29}=53 \).

Teacher's Note:
1. Use the standard formula and keep careful track of signs while subtracting the point vectors.
2. Verify the final value by an independent check where possible.

Answer: (a) (1, 0, 7)

Explanation:
1. Foot P = (t, 2t + 1, 3t + 2). The join vector (t - 1, 2t - 5, 3t - 1) is perpendicular to (1, 2, 3): \( t-1+4t-10+9t-3=14t-14=0 \), so t = 1.
2. P = (1, 3, 5).
3. Image = 2P - A = (1, 0, 7).

Teacher's Note:
1. The foot of the perpendicular is the mid-point of the point and its image.
2. So image = 2 x foot - point.