Let a=√7i+j-k and b=j+2k.

Question

Let $\vec{a}=\sqrt{7}\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r}	imes \vec{a}+\vec{a}	imes \vec{b}=\overset{	o }{0}$ and $\vec{r}\cdot \vec{a}=0$, then $|3\vec{r}{|}^{2}$ is equal to:

Accepted Answer

Given

$\vec{a}=\sqrt{7}\, \hat{i}+\hat{j}-\hat{k},\, \vec{b}=\hat{j}+2\hat{k}$

and

$\vec{r}	imes \vec{a}+\vec{a}	imes \vec{b}=\vec{0},\, \vec{r}\cdot \vec{a}=0$

We have to find $|3\vec{r}{|}^{2}$.

From

$\vec{r}	imes \vec{a}+\vec{a}	imes \vec{b}=\vec{0}$

note that

$\vec{a}	imes \vec{b}=-(\vec{b}	imes \vec{a})$

So,

$\vec{r}	imes \vec{a}=\vec{b}	imes \vec{a}$

Hence,

$(\vec{r}-\vec{b})	imes \vec{a}=0$

This means $\vec{r}-\vec{b}$ is parallel to $\vec{a}$. Therefore,

$\vec{r}=\vec{b}+\lambda \vec{a}$

for some scalar $\lambda $.

Now use the condition

$\vec{r}\cdot \vec{a}=0$

Substitute $\vec{r}=\vec{b}+\lambda \vec{a}$:

$(\vec{b}+\lambda \vec{a})\cdot \vec{a}=0$

So,

$\vec{b}\cdot \vec{a}+\lambda (\vec{a}\cdot \vec{a})=0$

Now calculate these dot products.

1. Find $\vec{b}\cdot \vec{a}$
$\vec{b}=(0,1,2),\, \vec{a}=(\sqrt{7},1,-1)$

Thus,

$\vec{b}\cdot \vec{a}=0\cdot \sqrt{7}+1\cdot 1+2\cdot (-1)=1-2=-1$

2. Find $\vec{a}\cdot \vec{a}$
$\vec{a}\cdot \vec{a}=(\sqrt{7}{)}^{2}+{1}^{2}+(-1{)}^{2}=7+1+1=9$

Therefore,

$-1+9\lambda =0$

So,

$\lambda =\frac{1}{9}$

Hence,

$\vec{r}=\vec{b}+\frac{1}{9}\vec{a}$

Now find $|\vec{r}{|}^{2}$.

Using

$\vec{r}=\vec{b}+\frac{1}{9}\vec{a}$

we get

$|\vec{r}{|}^{2}={|\vec{b}+\frac{1}{9}\vec{a}|}^{2}$

Expand:

$|\vec{r}{|}^{2}=|\vec{b}{|}^{2}+\frac{1}{81}|\vec{a}{|}^{2}+\frac{2}{9}(\vec{a}\cdot \vec{b})$

Now,

$|\vec{b}{|}^{2}={0}^{2}+{1}^{2}+{2}^{2}=5$

$|\vec{a}{|}^{2}=9$

$\vec{a}\cdot \vec{b}=-1$

So,

$|\vec{r}{|}^{2}=5+\frac{9}{81}+\frac{2}{9}(-1)$

$=5+\frac{1}{9}-\frac{2}{9}=5-\frac{1}{9}=\frac{44}{9}$

Therefore,

$|3\vec{r}{|}^{2}=9|\vec{r}{|}^{2}=9\cdot \frac{44}{9}=44$

So the correct answer is:

$44$

Hence, Option A.

Solutions · 1