A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i}+\hat{j}$ and the plane determined by the vectors $\hat{i}-\hat{j}, \hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ is :
Solution
<p>If ${\overrightarrow n _1}$ is a vector normal to the plane determined by $\widehat i$ and $\widehat i + \widehat j$ then</p>
<p>$${\overrightarrow n _1} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 0 & 0 \cr
1 & 1 & 0 \cr
} } \right| = \widehat k$$</p>
<p>If ${\overrightarrow n _2}$ is a vector normal to the plane determined by $\widehat i - \widehat j$ and $\widehat i + \widehat k$ then</p>
<p>$${\overline n _2} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 1} & 0 \cr
1 & 0 & 1 \cr
} } \right| = - \widehat i - \widehat j + \widehat k$$</p>
<p>Vector $\overrightarrow a$ is parallel to ${\overrightarrow n _1} \times {\overrightarrow n _2}$</p>
<p>i.e. $\overrightarrow a$ is parallel to $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 0 & 1 \cr
{ - 1} & { - 1} & 1 \cr
} } \right| = \widehat i - \widehat j$$</p>
<p>Given $\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k$</p>
<p>Cosine of acute angle between</p>
<p>$\overrightarrow a$ and $$\overrightarrow b = \left| {{{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow a |\,.\,|\overrightarrow b |}}} \right| = {1 \over {\sqrt 2 }}$$</p>
<p>Obtuse angle between $\overrightarrow a$ and $\overrightarrow b = {{3\pi } \over 4}$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.