Question

If $a=\frac{(3i+k)}{\sqrt{10}}\&b=\frac{(2i+3j-6k)}{7}$,then the value of $(2a-b).\left[\right(axb)\times (a+2b\left)\right]$is

• A. $-3$
• B. $5$
• C. $3$
• D. $-5$

Answer 1

The correct option is D

$-5$

Explanation for the correct option:

Finding cross product of vector:

Given, $a=\frac{(3i+k)}{\sqrt{10}}\&b=\frac{(2i+3j-6k)}{7}$

So,

$\begin{array}{rcl}& & (2a–b).\left[\right(axb)\times (a+2b\left)\right]\\ & =& (2a–b).\left[\right(axb)xa+(a\times b)\times 2b)]\\ & =& (2a–b).\left[\right(a.a)b–(b.a)a+2(a.b)b–2(b.b)a]\left[\because a\times (b\times c)=\left(a.c\right)b-\left(a.b\right)c\right]\\ \mathbf{\{}\mathbf{\because }\mathbf{a}\mathbf{.}\mathbf{a}& \mathbf{=}& \mathbf{b}\mathbf{.}\mathbf{b}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{=}\mathbf{0}\mathbf{\}}\end{array}$

Therefore,

$$\begin{gathered} =(2a–b)[1(b)–(0)a+2(0)b–2(1)a] \\ =(2a–b)(b–2a) \\ =–\left[4\right|a{|}^2+4a.b-\left|b{|}^2\right] \\ =–[4–0+1] \\ =-5 \end{gathered}$$

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}\mathbf{.}$