JEE Mathematics Straight Lines in Space MCQs Set 01

Answer: (b) \( \tfrac2{\sqrt{29}},\tfrac4{\sqrt{29}},\tfrac{-3}{\sqrt{29}} \)

Explanation:
1. Direction ratios are (2, 4, -3), with magnitude \( \sqrt{4+16+9}=\sqrt{29} \).
2. Divide each by \( \sqrt{29} \).

Teacher's Note:
1. Read the denominators as direction ratios.
2. Normalise them.

Answer: (a) \( \left(-\tfrac32,-\tfrac74,-\tfrac54\sqrt3\right) \)

Explanation:
1. Points on the line: \( (1+2t,\,2+3t,\,\sqrt3t) \). The given point is at t = -1.
2. The direction vector has length 4, so a point at distance 1 has parameter \( t=-1\pm\tfrac14 \).
3. t = -5/4 gives \( \left(-\tfrac32,-\tfrac74,-\tfrac{5\sqrt3}4\right) \).

Teacher's Note:
1. Parametrise the line.
2. Distance = |change in t| times |direction vector|.

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

Explanation:
1. Direction ratios are \( (\tfrac14,\tfrac13,0)\propto(3,4,0) \), through (1, -1, 2).
2. So the line is \( \dfrac{x-1}3=\dfrac{y+1}4=\dfrac{z-2}0 \).

Teacher's Note:
1. Eliminate r.
2. Scale the direction vector.

Answer: (a) \( \pi/6 \)

Explanation:
1. Direction vectors (1, -2, 1) and (-1, 1, 0): dot product = -3, magnitudes \( \sqrt6 \) and \( \sqrt2 \).
2. \( \cos\theta=\dfrac3{\sqrt{12}}=\dfrac{\sqrt3}2 \), so \( \theta=\pi/6 \).

Teacher's Note:
1. Use \( \cos\theta=|\vec a\cdot\vec b|/(|a||b|) \).
2. Take the acute angle.

Answer: (a) x = y = z

Explanation:
1. The required direction is the cross product of (-3, 2, 1) and (1, -3, 2).
2. \( (2\cdot2-1\cdot(-3),\ 1\cdot1-(-3)\cdot2,\ (-3)(-3)-2\cdot1)=(7,7,7) \), so the line is x = y = z.

Teacher's Note:
1. Take the cross product of the direction vectors.
2. Divide by 7.

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

Explanation:
1. Point (3, 0, 0) on the line, directions (3, 0, -4) and (0, 1, 0). Cross product = (4, 0, 3), magnitude 5.
2. Distance = \( \dfrac{|(3,0,0)\cdot(4,0,3)|}5=\dfrac{12}5 \).

Teacher's Note:
1. Use the formula for skew lines.
2. The y-axis passes through the origin.

Answer: (b) 6

Explanation:
1. Unit direction of the line is \( (\tfrac45,-\tfrac35,0) \).
2. Projection = \( 6\cdot\tfrac45+(-2)(-\tfrac35)+0=\tfrac{24+6}5=6 \).

Teacher's Note:
1. Dot the segment vector with the unit direction.
2. Normalise (4, -3, 0) first.

Answer: (b) ab

Explanation:
1. The vectors OA and OB have the same direction, so the angle between them is 0.
2. \( \vec{OA}\cdot\vec{OB}=ab\cos0=ab \).

Teacher's Note:
1. Interpret the sum as a dot product.
2. The angle between the vectors is zero.

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

Explanation:
1. The second direction \( (-1,\tfrac32,-3)=-\tfrac12(2,-3,6) \), so the lines are parallel.
2. Vector between points (4, -1, 0) and (0, 1, -1) is (4, -2, 1); cross with (2, -3, 6) gives (-9, -22, -8).
3. Distance = \( \dfrac{\sqrt{81+484+64}}{7}=\dfrac{\sqrt{629}}7 \).

Teacher's Note:
1. Check that the lines are parallel.
2. Use |cross product| / |direction|.

Answer: (a) \( \left(\tfrac{12-\sqrt3}4,-\tfrac52,\tfrac{13}4\right) \)

Explanation:
1. The direction vector \( (\sqrt3,2,3) \) has length 4, so the unit vector is \( (\tfrac{\sqrt3}4,\tfrac12,\tfrac34) \).
2. Moving in the opposite direction: \( (3-\tfrac{\sqrt3}4,\,-2-\tfrac12,\,4-\tfrac34)=\left(\tfrac{12-\sqrt3}4,-\tfrac52,\tfrac{13}4\right) \).

Teacher's Note:
1. Find the unit vector along the line.
2. Add or subtract it from A.

Answer: (c) \( \left(-3,\tfrac52,-\tfrac{\sqrt3}2\right) \)

Explanation:
1. Foot of the perpendicular \( (-1+2t,\,2+3t,\,\sqrt3t) \) satisfies \( 2(-1+2t)+3(2+3t)+3t=0 \), so \( 16t=-4 \) and \( t=-\tfrac14 \).
2. Foot = \( (-\tfrac32,\tfrac54,-\tfrac{\sqrt3}4) \); the image is twice this: \( (-3,\tfrac52,-\tfrac{\sqrt3}2) \).

Teacher's Note:
1. Find the foot of the perpendicular.
2. The image is twice the foot since the point is the origin.

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

Explanation:
1. A point is \( (1+t,\,2+2t,\,-1+3t) \); its squared distance is \( 14t^2+4t+6 \).
2. Setting this equal to 6: t = 0 or \( t=-\tfrac27 \). t = 0 gives (1, 2, -1).

Teacher's Note:
1. Parametrise and apply the distance condition.
2. Solve the resulting quadratic.