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 $$l^2+m^2+n^2=1$$
$$\begin{aligned} &\begin{aligned} & \Rightarrow \quad \frac{k^2}{49}+\frac{9 k^2}{196}+\frac{9 k^2}{49}=1 \\ & \Rightarrow \quad \frac{4 k^2+9 k^2+36 k^2}{196}=1 \\ & \Rightarrow \quad k^2=\frac{196}{49}=4 \\ & \Rightarrow \quad k= \pm 2 \end{aligned}\\ &\text { So, the direction cosines }(l, m, n) \text { are } \frac{2}{7}, \frac{3}{7} \text { and } \frac{-6}{7} \text {. } \end{aligned}$$
$$\begin{aligned} &\text { [since, } \vec{r} \text { makes an acute angle with } X \text {-axis] }\\ &\begin{array}{ll} \because \quad \overrightarrow{\mathbf{r}} & =\hat{\mathbf{r}} \cdot|\overrightarrow{\mathbf{r}}| \\ \therefore \quad \overrightarrow{\mathbf{r}} & =(l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}})|\overrightarrow{\mathbf{r}}| \\ & =\left(\frac{+2}{7} \hat{\mathbf{i}}+\frac{3}{7} \hat{\mathbf{j}}-\frac{6}{7} \hat{\mathbf{k}}\right) \cdot 14 \\ & =+4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{array} \end{aligned}$$
Find a vector of magnitude 6 , which is perpendicular to both the vectors $$ 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } 4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} .$$
Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
So, any vector perpendicular to both the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is given by
$$\begin{aligned} \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{array}\right| \\ & =\hat{\mathbf{i}}(-3+2)-\hat{\mathbf{j}}(6-8)+\hat{\mathbf{k}}(-2+4) \\ & =-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}=\overrightarrow{\mathbf{r}}\quad\text{[say]} \end{aligned}$$
$$\begin{aligned} &\text { A vector of magnitude } 6 \text { in the direction of } \vec{r}\\ &\begin{aligned} & =\frac{\overrightarrow{\mathbf{r}}}{|\overrightarrow{\mathbf{r}}|} \cdot 6=\frac{-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{1^2+2^2+2^2}} \cdot 6 \\ & =\frac{-6}{3} \hat{\mathbf{i}}+\frac{12}{3} \hat{\mathbf{j}}+\frac{12}{3} \hat{\mathbf{k}} \\ & =-2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned} \end{aligned}$$
Find the angle between the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}$.
Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
We know that, angle between two vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is given by
$$\begin{aligned} \cos \theta & =\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathbf{b}}|} \\ & =\frac{(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{4+1+1} \sqrt{9+16+1}} \\ & =\frac{6-4-1}{\sqrt{6} \sqrt{26}}=\frac{1}{2 \sqrt{39}} \\ \therefore\quad \theta & =\cos ^{-1}\left(\frac{1}{2 \sqrt{39}}\right) \end{aligned}$$
If $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0$, then show that $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$. Interpret the result geometrically.
$$\begin{array}{lrl} \text { Since, } & \vec{a}+\vec{b}+\vec{c} & =0 \\ \Rightarrow & \vec{b} & =-\vec{c}-\vec{a} \\ \text { Now, } & \vec{a} \times \vec{b} & =\vec{a} \times(-\vec{c}-\vec{a}) \end{array}$$
$=\vec{a} \times(-\vec{c})+\vec{a} \times(-\vec{a})=-\vec{a} \times \vec{c}$
$$\begin{array}{ll} \Rightarrow & \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \quad\text{.... (i)}\\ \text { Also, } & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathbf{c}}=(-\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}) \times \overrightarrow{\mathbf{c}} \\ & =(-\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}})+(-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}})=-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}} \\ \Rightarrow & \vec{b} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \quad\text{.... (ii)}\\ & \\ \text { From Eqs. (i) and (ii), } & \vec{a} \times \vec{b}=\vec{b} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \end{array}$$
Geometrical interpretation of the result
If $A B C D$ is a parallelogram such that $\overrightarrow{A B}=\overrightarrow{\mathbf{a}}$ and $\overrightarrow{A D}=\overrightarrow{\mathbf{b}}$ and these adjacent sides are making angle $\theta$ between each other, then we say that
Area of parallelogram $A B C D=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}||\sin \theta|=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
Since, parallelogram on the same base and between the same parallels are equal in area.
$$\begin{aligned} \text{We can say that,}\quad |\vec{a} \times \vec{b}| & =|\vec{a} \times \vec{c}|=|\vec{b} \times \vec{c}| \\ \text{This also implies that, }\quad\vec{a} \times \vec{b} & =\vec{a} \times \vec{c}=\vec{b} \times \vec{c} \end{aligned}$$
So, area of the parallelograms formed by taking any two sides represented by $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ as adjacent are equal.