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BITSAT 2005 Solved Question Paper

Section: Mathematics

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Question : 3 of 150

Marks: +1, -0

If

\vec{a}=\hat{i}+2\hat{j}-3\hat{k}

\vec{b}=3\hat{i}-\hat{j}+2\hat{k}

then the angle between the vectors

\vec{a}+\vec{b}

\vec{a}-\vec{b}

is:

$60^{\circ}$
$90^{\circ}$
$45^{\circ}$
$55^{\circ}$

Solution:

Given that

\vec{a}=\hat{i}+2\hat{j}-3\hat{k}

and

\vec{b}=3\hat{i}-\hat{j}+2\hat{k}

\vec{a}+\vec{b}=\hat{i}+2\hat{j}-3\hat{k}+3\hat{i}-\hat{j}+2\hat{k}

=4\hat{i}+\hat{j}-\hat{k}

and

\vec{a}-\vec{b}=(\hat{i}+2\hat{j}-3\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})

=-2\hat{i}+3\hat{j}-5\hat{k}

Let

\theta

be the angle between

\vec{a}+\vec{b}

and

\vec{a}-\vec{b}

\cos\theta=\frac{(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}|\,|\vec{a}-\vec{b}|}

=\frac{(4\hat{i}+\hat{j}-\hat{k})\cdot(-2\hat{i}+3\hat{j}-5\hat{k})}{|4\hat{i}+\hat{j}-\hat{k}|\,|-2\hat{i}+3\hat{j}-5\hat{k}|}

=\frac{-8+3+5}{\sqrt{16+1+1}\,\sqrt{4+9+25}}=0

\Rightarrow \theta = 90^{\circ}

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