Let a=2i+3j+4k,b=i-2j-2k and c=-i+4j+3k.

Question

Let $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k},\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$ and $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$.

If $\vec{d}$ is a vector perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{a}\cdot \vec{d}=18$, then $|\vec{a}	imes \vec{d}{|}^{2}$ is equal to :

Accepted Answer

Given vectors :

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

$\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$

$\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}$

Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, its direction is given by their cross product :

$\vec{d}=\lambda (\vec{b}	imes \vec{c})$

$\begin{gathered} \vec{b}	imes \vec{c} \quad  =|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  -2 \quad  -2 \ -1 \quad  4 \quad  3\end{array}| \ =\hat{i}(-6+8)-\hat{j}(3-2)+\hat{k}(4-2)=2\hat{i}-\hat{j}+2\hat{k} \end{gathered}$

Thus, 
$\vec{b}	imes \vec{c}=2\hat{i}-\hat{j}+2\hat{k}$

Given this, $\vec{d}$ can be expressed as :

$\vec{d}=\lambda (2\hat{i}-\hat{j}+2\hat{k})$

Using the given condition that $\vec{a}\cdot \vec{d}=18$ :

$(2\hat{i}+3\hat{j}+4\hat{k})\cdot \lambda (2\hat{i}-\hat{j}+2\hat{k})=18$

$\Rightarrow 2\lambda (2)-3\lambda (1)+4\lambda (2)=18$

$\Rightarrow \lambda (4+8-3)=18$

$\Rightarrow 9\lambda =18$

$\Rightarrow \lambda =2$

So, 
$\vec{d}=4\hat{i}-2\hat{j}+4\hat{k}$

Now, using the identity :

$|\vec{a}	imes \vec{d}{|}^{2}=|\vec{a}{|}^{2}|\vec{d}{|}^{2}-(\vec{a}\cdot \vec{d}{)}^{2}$

Given $|\vec{a}|=\sqrt{{2}^{2}+{3}^{2}+{4}^{2}}=\sqrt{29}$

and $|\vec{d}|=\sqrt{{4}^{2}+(-2{)}^{2}+{4}^{2}}=\sqrt{16+4+16}=\sqrt{36}=6$

Using your corrected calculations :

$|\vec{a}	imes \vec{d}{|}^{2}=29	imes 36-{18}^{2}=1044-324=720$

Solutions · 1