Question

If $b$and$c$ are any two non - collinear unit vectors and $a$ is any vector,

then $\frac{(a.b)b+(a.c)c+a.(bxc)}{|bxc|.(bxc)}$ is equal to

• A. $0$
• B. $a$
• C. $b$
• D. $c$

Answer 1

The correct option is B

$a$

Step 1. Find the value of $\frac{\mathbf{}\mathbf{(}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{)}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{.}\mathbf{c}\mathbf{)}\mathbf{c}\mathbf{}\mathbf{+}\mathbf{}\mathbf{a}\mathbf{.}\mathbf{(}\mathbf{b}\mathbf{}\mathbf{x}\mathbf{}\mathbf{c}\mathbf{)}}{\mathbf{|}\mathbf{b}\mathbf{}\mathbf{x}\mathbf{}\mathbf{c}\mathbf{|}\mathbf{.}\mathbf{(}\mathbf{b}\mathbf{}\mathbf{x}\mathbf{}\mathbf{c}\mathbf{)}}$:

Let $b=\hat{i},$

$c=\hat{j}$

$$\begin{gathered} \left|bxc\right|=\left|\hat{i}\times \hat{j}\right| \\ =\left|\hat{k}\right| \\ =1 \end{gathered}$$

Step 2. Suppose, vector $\mathit{a}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}\hat{\mathbf{i}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}\hat{\mathbf{j}}\mathbf{+}{\mathbf{a}}_{\mathbf{3}}\hat{\mathbf{k}}$

$$\begin{gathered} a.b=a.\hat{i} \\ =a_1, \end{gathered}$$

$$\begin{gathered} a.c=a.\hat{j} \\ =a_2 \end{gathered}$$

and $a.\frac{(b.c)}{\left|b\times c\right|}=a.\hat{k}$

$=a_3$

Step 3. Putting all the values in given equation we get;

$\left(a.b\right)b+\left(a.c\right)c+a.\frac{b.c}{\left|bxc\right|}\left(bxc\right)$

$=a_{1}b+a_{2}c+a_3\left(bxc\right)$

$=a_{1}i+a_{2}j+a_{3}K$

$=\mathit{a}$

Hence, option (B) is correct.