Let a=2i+3j+3k and b=6i+3j+3k.

Question

Let $\overset{	o }{a}=2\hat{i}+3\hat{j}+3\hat{k}$ and $\overset{	o }{b}=6\hat{i}+3\hat{j}+3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is :

Accepted Answer

Given

$\vec{a}=2\hat{i}+3\hat{j}+3\hat{k},\, \vec{b}=6\hat{i}+3\hat{j}+3\hat{k}$

We need the square of the area of the triangle whose adjacent sides are

$2\vec{a}+3\vec{b}\, \, \vec{a}-\vec{b}$

For two adjacent sides $\vec{u}$ and $\vec{v}$, area of triangle is

$\frac{1}{2}|\vec{u}	imes \vec{v}|$

So, square of the area is

${(\frac{1}{2}|\vec{u}	imes \vec{v}|)}^{2}=\frac{1}{4}|\vec{u}	imes \vec{v}{|}^{2}$

Let

$\vec{u}=2\vec{a}+3\vec{b},\, \vec{v}=\vec{a}-\vec{b}$

First find $\vec{u}$:

$2\vec{a}=2(2\hat{i}+3\hat{j}+3\hat{k})=4\hat{i}+6\hat{j}+6\hat{k}$

$3\vec{b}=3(6\hat{i}+3\hat{j}+3\hat{k})=18\hat{i}+9\hat{j}+9\hat{k}$

Therefore,

$\vec{u}=2\vec{a}+3\vec{b}=22\hat{i}+15\hat{j}+15\hat{k}$

Now find $\vec{v}$:

$\vec{v}=\vec{a}-\vec{b}=(2-6)\hat{i}+(3-3)\hat{j}+(3-3)\hat{k}=-4\hat{i}$

Now compute $\vec{u}	imes \vec{v}$:

$\vec{u}	imes \vec{v}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 22 \quad  15 \quad  15 \ -4 \quad  0 \quad  0 \end{vmatrix}$

Expanding,

$\vec{u}	imes \vec{v}=\hat{i}(15\cdot 0-15\cdot 0)-\hat{j}(22\cdot 0-15\cdot (-4))+\hat{k}(22\cdot 0-15\cdot (-4))$

$=0\hat{i}-60\hat{j}+60\hat{k}$

So,

$|\vec{u}	imes \vec{v}{|}^{2}={0}^{2}+(-60{)}^{2}+{60}^{2}=3600+3600=7200$

Hence square of the area of the triangle is

$\frac{1}{4}	imes 7200=1800$

Therefore, the correct answer is

$1800$

So, Option C is correct.

Solutions · 1