====== 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.

=====Question 6=====
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.
====Solution====
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}


=====Question 7=====
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 $\vec{b}$.
====Solution====
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}
$$\Rightarrow \quad \theta=\sin ^{-1}\left(\dfrac{\sqrt{3}}{2}\right)=60^\circ=\dfrac{\pi}{3}$$


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