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QJEEAdv10Ma0a38: JEE Advanced 2010 Physics - Class 11 JEEAdv10Ma0a38 Solution: Two adjacent sides of a parallelogram are given by and The side …

Two adjacent sides of a parallelogram ABCDABCD are given by
AB→=2i^+10j^+11k^\overrightarrow {AB} = 2\widehat i + 10\widehat j + 11\widehat k and  AD→=−i^+2j^+2k^\,\overrightarrow {AD} = -\widehat i + 2\widehat j + 2\widehat k
The side ADAD is rotated by an acute angle α\alpha in the plane of the parallelogram so that ADAD becomes AD′.AD'. If AD′AD' makes a right angle with the side AB,AB, then the cosine of the angle α\alpha is given by

Multiple Choice Options

  • A 89{{8 \over 9}}
  • B 179{{{\sqrt {17} } \over 9}}
  • C 19{{1 \over 9}}
  • D 459{{{4\sqrt 5 } \over 9}}

Step-by-Step Verified Solution

AB→=2i^+10j^+11k^\overrightarrow {AB} = 2\widehat i + 10\widehat j + 11\widehat k

IIT-JEE 2010 Paper 2 Offline Mathematics - Vector Algebra Question 62 English Explanation

AD→=−i^+2j^+2k^\overrightarrow {AD} = - \widehat i + 2\widehat j + 2\widehat k

Angle ' θ\theta ' between AB→\overrightarrow {AB} and AD→\overrightarrow {AD} is

cos⁡(θ)=∣AB→ . AD→∣AB→∣∣AD→∣∣\cos (\theta ) = \left| {{{\overrightarrow {AB} \,.\,\overrightarrow {AD} } \over {\left| {\overrightarrow {AB} } \right|\left| {\overrightarrow {AD} } \right|}}} \right|

=∣−2+20+22(15)(3)∣=89= \left| {{{ - 2 + 20 + 22} \over {(15)(3)}}} \right| = {8 \over 9}

⇒sin⁡(θ)=179\Rightarrow \sin (\theta ) = {{\sqrt {17} } \over 9}

Since, α+θ=90∘\alpha + \theta = 90^\circ

∴\therefore cos⁡(α)=cos⁡(90∘−θ)\cos (\alpha ) = \cos (90^\circ - \theta )

=sin⁡(θ)=179= \sin (\theta ) = {{\sqrt {17} } \over 9}

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