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Mathematics Question on Vector Algebra

A vector a⃗\vec{a}
is parallel to the line of intersection of the plane determined by the vectors
i^,i^+j^\hat{i},\hat{i}+\hat{j} and the plane determined by the vectors
i^−j^,i^+k^\hat{i}−\hat{j},\hat{i}+\hat{k}. The obtuse angle between
a⃗\vec{a} and the vector b⃗=i^−2j^+2k^\vec{b}=\hat{i}−2\hat{j}+2\hat{k}
is

A

3π4\frac{3π}{4}

B

2π3\frac{2π}{3}

C

4π5\frac{4π}{5}

D

5π6\frac{5π}{6}

Answer

3π4\frac{3π}{4}

Explanation

Solution

If n⃗1\vec{n}_1 is a vector normal to the plane determined by i^\hat{i} and i^+j^\hat{i} +\hat{j} then
n⃗1=\vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 1 & 0 & 0 \\\ 1 & 1 & 0 \\\ \end{vmatrix}$$= k
If n⃗2\vec{n}_2 is a vector normal to the plane determined by i^−j^\hat{i}-\hat{j} and i^+k^\hat{i} +\hat{k} then
n2 = ∣i^j^k^ 1−10 101 ∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 1 & -1 & 0 \\\ 1 & 0 & 1 \\\ \end{vmatrix}| = −i^−j^+k^-\hat{i}-\hat{j}+\hat{k}
Vector a⃗\vec{a} is parallel to n⃗1×n⃗2\vec{n}_1 \times \vec{n}_2 i.e
a⃗\vec{a} is parallel to∣i^j^k^ 001 −1−11 ∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 0 & 0 & 1 \\\ -1 & -1 & 1 \\\ \end{vmatrix} =i^−j^= \hat{i}-\hat{j}
Given
b=i^−2j^+2k^b=\hat{i}-2\hat{j}+2\hat{k}
cosine of acute angle between
a and b=∣a⃗.b⃗∣a⃗∣.∣b⃗∣∣=12a\ and\ b = |\frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}|=\frac{1}{\sqrt 2}
Obtuse angle between
a⃗\vec{a} and b⃗=3π4\vec{b} = \frac{3π}{4}