Skip to content
dronacharya.ai
← IIT-JEE 2010 · Paper 2
MathematicsSingle correct+5 / -2Question 24 of 57

IIT-JEE 2010 · Paper 2, Mathematics question 24

Two adjacent sides of a parallelogram ABCD are given by AB→=2i^+10j^+11k^\overrightarrow{AB}=2\hat i+10\hat j+11\hat k and AD→=−i^+2j^+2k^\overrightarrow{AD}=-\hat i+2\hat j+2\hat k.

The side AD is rotated by an acute angle α\alpha in the plane of the parallelogram so that AD becomes AD′. If AD′ makes a right angle with the side AB, then the cosine of the angle α\alpha is given by

Choose one option
Answer and worked solution

Official answer: B

Components of AD. ∣AD→∣=3|\overrightarrow{AD}|=3, ∣AB→∣=15|\overrightarrow{AB}|=15 and AD→⋅AB→=−2+20+22=40\overrightarrow{AD}\cdot\overrightarrow{AB}=-2+20+22=40. The component of AD along AB is 4015=83\frac{40}{15}=\frac83, so the component perpendicular to AB (in the plane) has length
9−649=173.\sqrt{9-\frac{64}9}=\frac{\sqrt{17}}3.

Rotated side. AD′ has length 3 and points along that perpendicular direction, so
cos⁡α=AD→⋅AD′→9=3⋅1739=179.\cos\alpha=\frac{\overrightarrow{AD}\cdot\overrightarrow{AD'}}{9}=\frac{3\cdot\frac{\sqrt{17}}3}{9}=\frac{\sqrt{17}}9.

(B)

Topics:Dot product
← Previous questionNext question →