Find a unit vector in the direction of $\overrightarrow{\mathbf{P Q}}$, where $P$ and $Q$ have coordinates $(5,0,8)$ and $(3,3,2)$, respectively.
Since, the coordinates of $P$ and $Q$ are $(5,0,8)$ and $(3,3,2)$, respectively.
$$\begin{aligned} \therefore \quad \overrightarrow{\mathrm{PQ}} & =\overrightarrow{\mathbf{O Q}}-\overrightarrow{\mathbf{O P}} \\ & =(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \\ & =-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ \therefore \quad \text { Unit vector in the direction of } \overrightarrow{\mathbf{P Q}} & =\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}=\frac{-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{\sqrt{2^2+3^2+6^2}} \\ & =\frac{-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{\sqrt{49}}=\frac{-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}}{7} \end{aligned}$$
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are the position vectors of $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ respectively, then find the position vector of a point $\overrightarrow{\mathbf{C}}$ in $\overrightarrow{\mathbf{B A}}$ produced such that $\overrightarrow{\mathbf{B C}}=1.5 \overrightarrow{\mathbf{B A}}$.
Since, $\overrightarrow{O A}=\vec{a}$ and $\overrightarrow{O B}=\vec{b}$
$$\begin{aligned} \therefore\quad\overrightarrow{\mathrm{BA}} & =\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \\ \text{and}\quad 1.5 \overrightarrow{\mathrm{BA}} & =1.5(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \\ \text{Since}\quad \overrightarrow{\mathrm{BC}} & =1.5 \overrightarrow{\mathrm{BA}}=1.5(\vec{a}-\vec{b}) \end{aligned}$$
$\overrightarrow{O C}-\overrightarrow{O B}=1.5 \vec{a}-1.5 \vec{b}$
$$\begin{aligned} \overrightarrow{\mathrm{OC}} & =1.5 \overrightarrow{\mathrm{a}}-1.5 \overrightarrow{\mathrm{~b}}+\overrightarrow{\mathrm{b}} \quad [\because \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}]\\ & =1.5 \overrightarrow{\mathrm{a}}-0.5 \overrightarrow{\mathrm{~b}} \\ & =\frac{3 \vec{a}-\vec{b}}{2} \end{aligned}$$
Graphically, explanation of the above solution is given below
Using vectors, find the value of $k$, such that the points $(k,-10,3)$, $(1,-1,3)$ and $(3,5,3)$ are collinear.
Let the points are $A(k,-10,3), B(1,-1,3)$ and $C(3,5,3)$.
$$\begin{aligned} \text{So,}\quad \overrightarrow{A B} & =\overrightarrow{O B}-\overrightarrow{\mathbf{O A}} \\ & =(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(k \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & =(1-k) \hat{\mathbf{i}}+(-1+10) \hat{\mathbf{j}}+(3-3) \hat{\mathbf{k}} \\ & =(1-k) \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+0 \hat{\mathbf{k}} \\ \therefore\quad |\overrightarrow{\mathbf{A B}}| & =\sqrt{(1-k)^2+(9)^2+0}=\sqrt{(1-k)^2+81} \end{aligned}$$
$$\begin{aligned} &\text { Similarly, }\\ &\begin{aligned} \overrightarrow{B C} & =\overrightarrow{O C}-\overrightarrow{O B} \\ & =(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+0 \hat{\mathbf{k}} \\ \therefore\quad|\overrightarrow{B C}| & =\sqrt{2^2+6^2+0}=2 \sqrt{10} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text{and}\quad \overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\ & =(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(k \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & =(3-k) \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+0 \hat{\mathbf{k}} \\ \therefore\quad |\overrightarrow{\mathbf{A C}}| & =\sqrt{(3-k)^2+225} \end{aligned}$$
If $A, B$ and $C$ are collinear, then sum of modulus of any two vectors will be equal to the modulus of third vectors
$$\begin{aligned} &\begin{array}{l} \text { For }|\overrightarrow{\mathbf{A B}}|+|\overrightarrow{\mathbf{C C}}|=|\overrightarrow{\mathbf{A C}}|, \\ \Rightarrow \quad \sqrt{(1-k)^2+81}+2 \sqrt{10}=\sqrt{(3-k)^2+225} \\ \Rightarrow \quad \sqrt{(3-k)^2+225}-\sqrt{(1-k)^2+81}=2 \sqrt{10} \\ \Rightarrow \quad \sqrt{9+k^2-6 k+225}-\sqrt{1+k^2-2 k+81}=2 \sqrt{10} \\ \Rightarrow \quad \sqrt{k^2-6 k+234}-2 \sqrt{10}=\sqrt{k^2-2 k+82} \\ \Rightarrow \quad k^2-6 k+234+40-2 \sqrt{k^2-6 k+234} \cdot 2 \sqrt{10}=k^2-2 k+82 \\ \Rightarrow \quad k^2-6 k+234+40-k^2+2 k-82=4 \sqrt{10} \sqrt{k^2+234-6 k} \\ \Rightarrow \quad -4 k+192=4 \sqrt{10} \sqrt{k^2+234-6 k} \\ \Rightarrow \quad -k+48=\sqrt{10} \sqrt{k^2+234-6 k} \end{array}\\ &\text { On squaring both sides, we get }\\ &\begin{array}{rlrl} & 48 \times 48+k^2-96 k =10\left(k^2+234-6 k\right) \\ \Rightarrow & k^2-96 k-10 k^2+60 k =-48 \times 48+2340 \\ \Rightarrow & -9 k^2-36 k =-48 \times 48+2340 \end{array} \end{aligned}$$
$$\begin{array}{lrl} \Rightarrow & \left(k^2+4 k\right) & =+16 \times 16-260 \quad \text{[dividing by 9 in both sides]}\\ \Rightarrow & k^2+4 k & =-4 \\ & k^2+4 k+4 & =0 \\ \Rightarrow & (k+2)^2 & =0 \\ \therefore & k & =-2 \end{array}$$
A vector $\overrightarrow{\mathbf{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\overrightarrow{\mathbf{r}}$ is $2 \sqrt{3}$ units, then find the value of $\overrightarrow{\mathbf{r}}$.
We have, $$|\overrightarrow{\mathbf{r}}|=2 \sqrt{3}$$
Since, $\vec{r}$ is equally inclined to the three axes, $\vec{r}$ so direction cosines of the unit vector $\vec{r}$ will be same. i.e., $l=m=n$.
We know that,
$$\begin{array}{r} & l^2+m^2+n^2 =1 \\ \Rightarrow & l^2+l^2+l^2 =1 \\ \Rightarrow & l^2 =\frac{1}{3} \end{array}$$
$$\begin{array}{ll} \Rightarrow & l= \pm\left(\frac{1}{\sqrt{3}}\right) \\ \text { So, } & \hat{\mathbf{r}}= \pm \frac{1}{\sqrt{3}} \hat{\mathbf{i}} \pm \frac{1}{\sqrt{3}} \hat{\mathbf{j}} \pm \frac{1}{\sqrt{3}} \hat{\mathbf{k}} \\ \therefore & \overrightarrow{\mathbf{r}}=\hat{\mathbf{r}} \mid \overrightarrow{\mathbf{r}}\quad \left[\because \hat{r}=\frac{\vec{r}}{|\vec{r}|}\right] \end{array}$$
$$\begin{aligned} & =\left[ \pm \frac{1}{\sqrt{3}} \hat{\mathbf{i}} \pm \frac{1}{\sqrt{3}} \hat{\mathbf{j}} \pm \frac{1}{\sqrt{3}} \hat{\mathbf{k}}\right] 2 \sqrt{3} \quad[\because|r|=2 \sqrt{3}] \\ & = \pm 2 \hat{\mathbf{i}} \pm 2 \hat{\mathbf{j}} \pm 2 \hat{\mathbf{k}}= \pm 2(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned}$$
If a vector $\overrightarrow{\mathbf{r}}$ has magnitude 14 and direction ratios 2,3 and -6 . Then, find the direction cosines and components of $\overrightarrow{\mathbf{r}}$, given that $\overrightarrow{\mathbf{r}}$ makes an acute angle with $X$-axis.
Here, $|\overrightarrow{\mathbf{r}}|=14, \overrightarrow{\mathbf{a}}=2 k, \overrightarrow{\mathbf{b}}=3 k$ and $\overrightarrow{\mathbf{c}}=-6 k$
$\therefore$ Direction cosines $l, m$ and $n$ are
$$\begin{aligned} & l=\frac{\overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{r}}|}=\frac{2 k}{14}=\frac{k}{7} \\ & m=\frac{\overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{r}}|}=\frac{3 k}{14} \\ \text{and}\quad & n=\frac{\overrightarrow{\mathbf{c}}}{|\overrightarrow{\mathbf{r}}|}=\frac{-6 k}{14}=\frac{-3 k}{7} \end{aligned}$$
Also, we know that $$l^2+m^2+n^2=1$$
$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad \frac{k^2}{49}+\frac{9 k^2}{196}+\frac{9 k^2}{49}=1 \\ & \Rightarrow \quad \frac{4 k^2+9 k^2+36 k^2}{196}=1 \\ & \Rightarrow \quad k^2=\frac{196}{49}=4 \\ & \Rightarrow \quad k= \pm 2 \end{aligned}\\ &\text { So, the direction cosines }(l, m, n) \text { are } \frac{2}{7}, \frac{3}{7} \text { and } \frac{-6}{7} \text {. } \end{aligned}$$
$$\begin{aligned} &\text { [since, } \vec{r} \text { makes an acute angle with } X \text {-axis] }\\ &\begin{array}{ll} \because \quad \overrightarrow{\mathbf{r}} & =\hat{\mathbf{r}} \cdot|\overrightarrow{\mathbf{r}}| \\ \therefore \quad \overrightarrow{\mathbf{r}} & =(l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}})|\overrightarrow{\mathbf{r}}| \\ & =\left(\frac{+2}{7} \hat{\mathbf{i}}+\frac{3}{7} \hat{\mathbf{j}}-\frac{6}{7} \hat{\mathbf{k}}\right) \cdot 14 \\ & =+4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{array} \end{aligned}$$