Question

If $a=2\stackrel{\wedge }{i}-\stackrel{\wedge }{j}-m\stackrel{\wedge }{k}\text{ and }b={\displaystyle \frac{4}{7}}\stackrel{\wedge }{i}-{\displaystyle \frac{2}{7}}\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}$ are collinear, then the value of m is equal to
(A) -7
(B) -1
(C) 2
(D) 7
(E) -2

Answer 1

Hint: Two vectors are collinear if they are in the same line. Use formula $\overrightarrow{A}.\overrightarrow{B}=\left(\overrightarrow{A}\right)\left(\overrightarrow{B}\right)\cos \theta$ where $\theta$is angle between $\overrightarrow{A}\text{ and }\overrightarrow{B}$.

Complete step-by-step answer:
Here, we have two vectors given as $\overrightarrow{a}=2i-j-mk\text{ and }\overrightarrow{b}={\displaystyle \frac{4}{7}}i-{\displaystyle \frac{2}{7}}j+2k.$
Now, we have to find value of m if $\overrightarrow{a}\text{ and }\overrightarrow{b}$ are collinear.
As, we know collinear means in a line, that is if two vectors $\overrightarrow{a}\text{ and }\overrightarrow{b}$ are collinear then they lie in the same line.
In another language, we can say that if two vectors are collinear then the angle between them is $0{}^{\circ }$.
Hence, $\overrightarrow{a}\text{ and }\overrightarrow{b}$ have angle $0{}^{\circ }$ between them.
Now, we know dot product of two vectors can be given by;
$\overrightarrow{A}.\overrightarrow{B}=\left(\overrightarrow{A}\right)\left(\overrightarrow{B}\right)\cos \theta$
Where $\theta$ is angle between $\overrightarrow{A}\text{ and }\overrightarrow{B}$.
As, we have two vectors $\overrightarrow{a}=2i-j-mk\text{ and }\overrightarrow{b}={\displaystyle \frac{4}{7}}i-{\displaystyle \frac{2}{7}}j+2\stackrel{\wedge }{k}$ which have $0{}^{\circ }$ between them.
Hence,
$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$………………….(1)
We know magnitude of any vector $\overrightarrow{r}=xi-yj-zk$ is given as;
$\left|\overrightarrow{r}\right|=\sqrt{x^2+y^2+z^2}$
Hence, equation (1) can be written as
We know that $\cos 0{}^{\circ }=1$ and can simplify the above equation as;
$\begin{array}{rl}& \overrightarrow{a}.\overrightarrow{b}=\sqrt{5+m^2}\sqrt{{\displaystyle \frac{20}{49}}+4}\left(1\right)\\ & \overrightarrow{a}.\overrightarrow{b}=\sqrt{5+m^2}{\displaystyle \frac{\sqrt{216}}{7}}\\ & or\\ & \overrightarrow{a}.\overrightarrow{b}={\displaystyle \frac{6\sqrt{6}}{7}}\sqrt{5+m^2}.............\left(2\right)\end{array}$
Now, we can rewrite $\overrightarrow{a}.\overrightarrow{b}$ in different way as;
$\overrightarrow{a}.\overrightarrow{b}=(2i-j-mk).({\displaystyle \frac{4}{7}}i-{\displaystyle \frac{2}{7}}j+2k)............\left(3\right)$
As we know the same vectors have angle $0{}^{\circ }$. Hence, angles between (i, j), (j, k) and (i, k) are perpendicular to each other as they are unit vectors through the x, y and z axis.
Hence, equation (3) can be solved as;
$\begin{array}{rl}& \overrightarrow{a}.\overrightarrow{b}=2\times {\displaystyle \frac{4}{7}}+1\times {\displaystyle \frac{2}{7}}-m\times 2\\ & \overrightarrow{a}.\overrightarrow{b}={\displaystyle \frac{10}{7}}-2m..............\left(4\right)\end{array}$
Now, equation (2) and (4) have equal L.H.S hence R.H.S should also be equal to each other.
$\overrightarrow{a}.\overrightarrow{b}={\displaystyle \frac{6\sqrt{6}}{7}}\sqrt{5+m^2}={\displaystyle \frac{10}{7}}-2m$
Squaring both sides of the equation we get;
$\begin{array}{rl}& {\displaystyle \frac{216(5+m^2)}{49}}={\displaystyle \frac{{(10-14m)}^2}{49}}\\ & 216\times 5+216m^2=100+196m^2-280m\\ & 20m^2+280m+980=0\end{array}$
Dividing the whole equation by 20, we get;
$m^2+14m+49=0$
Now, we can factorize above equation as;
${(m+7)}^2=0$
Hence,
$\begin{array}{rl}& m+7=0\\ & m=-7\end{array}$
Therefore, for m = -7, $\overrightarrow{a}\text{ and }\overrightarrow{b}$ as given in question are collinear.
Note: One can go wrong while putting angle between two vectors $\overrightarrow{a}\text{ and }\overrightarrow{b}$which is $0{}^{\circ }$. He/she may put the angle be $180{}^{\circ }$ which is wrong.
Need to be careful with angles between $\stackrel{\wedge }{i},\stackrel{\wedge }{j}$and $\stackrel{\wedge }{k}$ during evaluation of $\overrightarrow{a}\text{ and }\overrightarrow{b}$.