Question 6 & 7 Review Exercise 3

Solutions of Question 6 & 7 of Review Exercise 3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find $\lambda$. if the veclors $\vec{a}=\hat{i}+3 \hat{j}+\hat{k}$, $\bar{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{j}+3 \hat{k}$ are coplanar.

Since the given vectors are coplanar, therefore,

\begin{align}\vec{a} \cdot \vec{b} \times \vec{c}&=0 \\ \Rightarrow\left|\begin{array}{ccc} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{array}\right|&=0 \\ \Rightarrow \quad 1(-3+\lambda)-3(6+0)+1(2 \lambda-0)&=0\\ \Rightarrow \quad-3+\lambda-18+2 \lambda&=0 \\ \Rightarrow \quad 3 \lambda - 21&=0 \\ \Rightarrow \quad \lambda&=\dfrac{21}{3}=7 .\end{align}

Vector $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=\sqrt{3}$, and $|\vec{b}|=\dfrac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector. Write the angle between $\vec{a}$ and $\vec{b}$.

Let $\theta$ be the angle between two vectors. We are given
$$|\vec{a} \times \vec{b}|=1,|\vec{a}|=\sqrt{3} \text { and }|\vec{c}|=\dfrac{2}{3} \text {. }$$ We know that $$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$$. Putting the given in above, we get \begin{align}1&=\sqrt{3} \cdot \dfrac{2}{3} \sin \theta \\ \Rightarrow \sqrt{3} \cdot \dfrac{2}{\sqrt{3} \sqrt{3}} \sin \theta&=1 \\ \Rightarrow \sin \theta&=\dfrac{\sqrt{3}}{2}\end{align} $$\Rightarrow \quad \theta=\sin ^{-1}\left(\dfrac{\sqrt{3}}{2}\right)=60^\circ=\dfrac{\pi}{3}$$