JEE Mathematics Three Dimensional Geometry MCQs Set 18

Answer: (b) \( \frac{2}{\sqrt{29}},\frac{4}{\sqrt{29}},\frac{-3}{\sqrt{29}} \)

Explanation:
1. The direction ratios are 2, 4, −3.
2. Their magnitude is \( \sqrt{4+16+9}=\sqrt{29} \).
3. Direction cosines = ratios divided by √29.

Teacher's Note:
1. Direction cosines are direction ratios divided by the square root of the sum of their squares.
2. The sum of squares of direction cosines is always 1.

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

Explanation:
1. The line has direction ratios 3 : 2 : 1, so options (a) and (b) (ratios 1 : 2 : 3) are wrong.
2. In option (c), the point (1, 2, 3) gives (1 + 2)/3 = 1, 2/2 = 1 and (3 − 2)/1 = 1, so the point lies on the line.
3. Hence (c) is the line required.

Teacher's Note:
1. Check both the direction ratios and that the given point satisfies the equations.
2. Different forms can represent the same line, since any point of the line may be used.

Answer: (a) \( \frac{x+7}{5}=\frac{y+2}{3}=\frac{z-1}{-1} \)

Explanation:
1. Direction ratios = (3 + 2, 4 − 1, −1 − 0) = (5, 3, −1).
2. Option (a) has these ratios and passes through (−2, 1, 0): (−2 + 7)/5 = 1, (1 + 2)/3 = 1, (0 − 1)/(−1) = 1.
3. Hence (a) is correct.

Teacher's Note:
1. The two-point form uses the difference of coordinates as direction ratios.
2. Verify an option by substituting one of the given points.

Answer: (a) \( \left(-\frac{3}{2},-\frac{7}{4},-\frac{5}{4}\sqrt{3}\right) \)

Explanation:
1. The given point is on the line: t = −1 gives (−1, −1, −√3).
2. The direction vector (2, 3, √3) has length 4, so the unit vector is (2, 3, √3)/4.
3. Points at distance 1: (−1, −1, −√3) ± (1/2, 3/4, √3/4).
4. The minus sign gives (−3/2, −7/4, −5√3/4), which is option (a).

Teacher's Note:
1. Move a distance d along a line by adding d times the unit direction vector.
2. There are always two such points, one on each side.

Answer: (b) \( \frac{x-1}{3}=\frac{y+1}{4}=\frac{z-2}{0} \)

Explanation:
1. The point is (1, −1, 2) + r(1/4, 1/3, 0), a line with direction ratios 1/4 : 1/3 : 0 = 3 : 4 : 0.
2. So the line is (x − 1)/3 = (y + 1)/4 = (z − 2)/0.

Teacher's Note:
1. Write the moving point as a fixed point plus a parameter times a direction vector.
2. Scale the direction vector to integers.

Answer: (c) \( \left(2,\frac{1}{3}\right) \)

Explanation:
1. Parallel lines have proportional direction ratios: (4λ + 1)/(−3) = 4/(5μ − 3) = −18/6 = −3.
2. 4λ + 1 = 9 gives λ = 2.
3. 4/(5μ − 3) = −3 gives 5μ − 3 = −4/3, so μ = 1/3.

Teacher's Note:
1. Use the third ratio, which has no unknown, to get the common proportionality factor.
2. Then solve for each unknown separately.

Answer: (a) \( \frac{\pi}{6} \)

Explanation:
1. Direction vectors are (1, −2, 1) and (−1, 1, 0).
2. The dot product is −1 − 2 = −3; the lengths are √6 and √2.
3. |cos θ| = 3/√12 = √3/2, so the acute angle is π/6.

Teacher's Note:
1. The angle between lines is the acute angle between their direction vectors, so take the modulus of the cosine.
2. cos θ = √3/2 corresponds to 30°.

Answer: (b) (α + α′, β + β′, γ + γ′)

Explanation:
1. The point (α + α′, β + β′, γ + γ′) satisfies the first line with common ratio 1, because each fraction becomes α/α, β/β, γ/γ.
2. It satisfies the second line with common ratio 1 as well.
3. Since the ratios are not all equal the lines are not parallel, so this is their point of intersection.

Teacher's Note:
1. Test a candidate point in both equations rather than solving from scratch.
2. A common ratio of 1 for both lines is the quick signal here.

Answer: (a) x = y = z

Explanation:
1. The required direction is the cross product of (−3, 2, 1) and (1, −3, 2).
2. Cross product = (2·2 − 1·(−3), 1·1 − (−3)·2, (−3)(−3) − 2·1) = (7, 7, 7).
3. A line through the origin with ratios 1 : 1 : 1 is x = y = z.

Teacher's Note:
1. A line perpendicular to two lines has direction equal to the cross product of their directions.
2. Simplify the cross product by dividing out a common factor.

Answer: (d) \( \frac{12}{5} \)

Explanation:
1. The line passes through (3, 0, 0) with direction (3, 0, −4); the y-axis passes through the origin with direction (0, 1, 0).
2. The cross product of the directions is (4, 0, 3), of length 5.
3. The vector between the points (3, 0, 0) dotted with (4, 0, 3) is 12.
4. Shortest distance = 12/5.

Teacher's Note:
1. Shortest distance between skew lines = |(P₂ − P₁)·(d₁ × d₂)| / |d₁ × d₂|.
2. A zero direction ratio means the line is parallel to a coordinate plane.

Answer: (b) 6

Explanation:
1. The segment vector is (6, −2, 1) and the line direction is (4, −3, 0), of length 5.
2. Projection = |(6)(4) + (−2)(−3) + 0| / 5 = 30/5 = 6.

Teacher's Note:
1. Projection of a vector on a line is the dot product with the unit direction vector.
2. The position of the line does not matter, only its direction.

Answer: (b) ab

Explanation:
1. Both points lie along the same ray from the origin, so the angle between OA and OB is 0.
2. OA·OB = pp′ + qq′ + rr′ = |OA||OB| cos 0 = ab.

Teacher's Note:
1. The dot product equals product of lengths times the cosine of the angle between the vectors.
2. If the points were on opposite sides of the origin the result would be −ab.

Answer: (d) infinite

Explanation:
1. The lines pass through (k, 1, −1) and (k + 1, 0, 1); the vector joining them is (1, −1, 2).
2. This vector equals the direction (1, −1, 2) of the second line, so the scalar triple product of the joining vector and the two directions is zero for every k.
3. The directions (4, 2, 1) and (1, −1, 2) are not proportional, so the lines are not parallel; hence they intersect for every real k.

Teacher's Note:
1. Lines intersect if they are coplanar and not parallel.
2. When the joining vector repeats one of the directions, the determinant has two equal rows.

Answer: (c) \( \frac{\sqrt{629}}{7} \)

Explanation:
1. The directions (2, −3, 6) and (−1, 3/2, −3) are proportional, so the lines are parallel.
2. Take P₁ = (4, −1, 0) and P₂ = (0, 1, −1); P₂ − P₁ = (−4, 2, −1).
3. (P₂ − P₁) × (2, −3, 6) = (9, 22, 8), of length √629; the direction has length 7.
4. Distance = √629 / 7.

Teacher's Note:
1. For parallel lines, distance = |(P₂ − P₁) × d| / |d|.
2. Check parallelism first by comparing direction ratios.

Answer: (a) \( \left(\frac{12-\sqrt{3}}{4},-\frac{5}{2},\frac{13}{4}\right) \)

Explanation:
1. The direction (√3, 2, 3) has length 4, so the unit vector is (√3/4, 1/2, 3/4).
2. Moving by −1 along it: (3 − √3/4, −2 − 1/2, 4 − 3/4) = ((12 − √3)/4, −5/2, 13/4).

Teacher's Note:
1. Shifting parallel to a line means adding a multiple of its unit direction vector.
2. Both ± directions are valid, but only one appears in the options.

Answer: (c) \( \left(-3,\frac{5}{2},-\frac{\sqrt{3}}{2}\right) \)

Explanation:
1. A general point is (−1 + 2t, 2 + 3t, √3 t). The foot of the perpendicular from the origin satisfies 2(−1 + 2t) + 3(2 + 3t) + 3t = 0, i.e. 4 + 16t = 0, so t = −1/4.
2. The foot is (−3/2, 5/4, −√3/4).
3. The image is twice the foot: (−3, 5/2, −√3/2).

Teacher's Note:
1. The image of a point in a line is the reflection through the foot of the perpendicular, so it equals twice the foot (for the origin).
2. Find the foot by making the join perpendicular to the direction.

Answer: (a) \( \frac{4}{3} \)

Explanation:
1. The line passes through the origin with direction (3, 0, 4), of length 5.
2. (1, 2, λ) × (3, 0, 4) = (8, 3λ − 4, −6).
3. Distance² = [64 + (3λ − 4)² + 36]/25 = 4, so (3λ − 4)² = 0.
4. Hence λ = 4/3.

Teacher's Note:
1. Distance of a point from a line = |cross product of the join and the direction| / |direction|.
2. A perfect square equal to zero gives a unique value.

Answer: (d) h = 2, k = 1/2

Explanation:
1. The first two lines meet at (1, 2, 3): t = 1 on the first and s = 0 on the second.
2. The third line must pass through (1, 2, 3): (1 + k)/3 = (2 − 1)/2 = 1/2 gives k = 1/2.
3. (3 − 2)/h = 1/2 gives h = 2.

Teacher's Note:
1. Concurrent lines pass through one common point; find it from two lines and force the third through it.
2. Equate the common ratio to the known fraction first.

Answer: (a) 2

Explanation:
1. The points (1, −1, 0) and (0, k, 1) give the joining vector (−1, k + 1, 1); the directions are (4, 3, k) and (1, 3, −2).
2. Coplanarity requires the determinant of these three vectors to vanish: k² + 12k + 23 = 0.
3. The discriminant 144 − 92 = 52 is positive, so there are 2 real values of k.

Teacher's Note:
1. Coplanar lines have a zero scalar triple product of the joining vector and the two directions.
2. Count real roots with the discriminant.

Answer: (b) 2 and 3

Explanation:
1. A point of the line is (1 + t, 2 + 2t, −1 + 3t). Setting its squared distance from the origin equal to 6 gives 14t² + 4t = 0.
2. t = 0 gives (1, 2, −1) and t = −2/7 gives (5/7, 10/7, −13/7).
3. So statements 2 and 3 are correct.

Teacher's Note:
1. Substitute the parametric form into the distance condition and solve the quadratic.
2. A line meets a sphere centred at the origin in at most two points.

Answer: (c) 1 and 4

Explanation:
1. For direction cosines (l, m, n) the angle condition is |l + 2m − n|/√6 = √(6/11), so |l + 2m − n| = 6/√11.
2. Statements 1, 2 and 4 satisfy this; statement 3 gives 2/√6, which fails.
3. The line must also meet the given line, so (l, m, n) must be coplanar with the point (−2, 1, 0) and the direction (1, 2, −1): the determinant is 0 for statements 1 and 4 and non-zero for statement 2.
4. So statements 1 and 4 are correct.

Teacher's Note:
1. Two conditions are needed: the angle and the intersection (coplanarity).
2. Check each candidate against both conditions.