Question 9 & 10, Exercise 3.2

Solutions of Question 9 & 10 of Exercise 3.2 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.

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, $$\overrightarrow{b}=4\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$, find a vector of magnitude of $6$ unit which is parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$

We compute that
\begin{align}2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}&=2(\hat{i}+\hat{j}+\hat{k})-(4\hat{i}-2\hat{j}+3\hat{k})+3(\hat{i}-2\hat{j}+\hat{k})\\ &=(2-4+3)\hat{i}-(2+2-6)\hat{j}+(2-3+3)\hat{k}\\ 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}&=\hat{i}-2\hat{j}+2\hat{k}\\ |2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}|&=\sqrt{(1)^2+(-2)^2+(2)^2}\\ &=\sqrt{9}=3\end{align} Let sat $\hat{a}$ be unit vector in direction of $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$ \begin{align}\hat{a}&=\dfrac{2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}}{|2\overrightarrow{a}\overrightarrow{b}+3\overrightarrow{c}|}\\ &=\dfrac{\hat{i}-2\hat{j}+2\hat{k}}{3}\end{align} Now vector of magnitude of $6$ unit which is parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$is
\begin{align}6\hat{a}&=6[\frac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})]\\ &=2(\hat{i}-2\hat{j}+2\hat{k})\\ \implies 6\hat{a}&=2\hat{i}-4\hat{j}+4\hat{k}\end{align} This is a desired vector with given conditions.

Find the position vector of a point $R$ which divides the line joining the points whose position vectors are $P(\hat{i}+2\hat{j}-\hat{k})$and $Q(-\hat{i}+\hat{j}+\hat{k})$ in the ratio $2:1$ internally and externally.

We find the position vector of a point $R$ which divides the line joining the points internally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\cdot \overrightarrow{OP}+2\cdot \overrightarrow{OQ}}{1+2}\\ &=\dfrac{1(\hat{i}+2\hat{j}-\hat{k})+2(-\hat{i}+\hat{j}+\hat{k})}{3}\\ &=\dfrac{-\hat{i}+4\hat{j}+\hat{k}}{3}\\ \implies \overrightarrow{OR}&=-\dfrac{1}{3}\hat{i}+\dfrac{4}{3}\hat{j}+\dfrac{1}{3}\hat{k}\end{align} Now we find the position vector of a point $\vec{R}$, which divides the line joining the points externally. By using ratio theorem, we have \begin{align}\overrightarrow{OR}&=\dfrac{1\cdot \overrightarrow{OP}-2\cdot \overrightarrow{OQ}}{1-2}\\ &=\dfrac{1(\hat{i}+2\hat{j}-\hat{k})-2(-\hat{i}+\hat{j}+\hat{k})}{-1}\\ &=-\hat{i}-2\hat{j}+\hat{k}-2\hat{i}+2\hat{j}+2\hat{k}\\ \implies\quad \text{externally}\quad \overrightarrow{OR}&=-3\hat{i}+3\hat{k}\end{align}

< Question 8 Question 11 >