Question

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

• A. \(\frac{3π}{4}\)
• B. \(\frac{2π}{3}\)
• C. \(\frac{4π}{5}\)
• D. \(\frac{5π}{6}\)

Answer 1

The Correct Option is A

If \(\vec{n}_1\) is a vector normal to the plane determined by \(\hat{i} \) and \(\hat{i} +\hat{j}\) then
\(\vec{n}_1 =\) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \end{vmatrix}\)\(= k\)
If \(\vec{n}_2\) is a vector normal to the plane determined by \(\hat{i}-\hat{j}\) and \(\hat{i} +\hat{k}\) then
n2 = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1 \\ \end{vmatrix}\)| = \(-\hat{i}-\hat{j}+\hat{k}\)
Vector \(\vec{a}\) is parallel to \(\vec{n}_1 \times \vec{n}_2\) i.e
\(\vec{a}\) is parallel to\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ -1 & -1 & 1 \\ \end{vmatrix}\) \(= \hat{i}-\hat{j}\)
Given 
\(b=\hat{i}-2\hat{j}+2\hat{k}\)
cosine of acute angle between 
\(a\ and\ b = |\frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}|=\frac{1}{\sqrt 2}\)
Obtuse angle between
\(\vec{a} \) and \(\vec{b} = \frac{3π}{4}\)