Let a=i+j+k,b=2i+4j-5k and c=xi+2j+3k,x R.

Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}$ and $\vec{c}=x\hat{i}+2\hat{j}+3\hat{k},x\in R$.
If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a}\cdot \vec{d}=1$, then $(\vec{a}	imes \vec{b})\cdot \vec{c}$ is equal to

Accepted Answer

$\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}+4\hat{j}-5\hat{k},\vec{c}=x\hat{i}+2\hat{j}+3\hat{k},x\in R$<?p>
$\vec{b}+\vec{c}=(x+2)\hat{i}+6\hat{j}-2\hat{k}$

$\vec{d}\vec{b}+\vec{c}$

$\begin{gathered} |\vec{b}+\vec{c}|=\sqrt{(x+2{)}^{2}+{6}^{2}+{2}^{2}} \ =\sqrt{40+(x+2{)}^{2}} \ \vec{d}=\frac{x+2}{\sqrt{40+(x+2{)}^{2}}}\hat{i}+\frac{6}{\sqrt{40+(x+2{)}^{2}}}\hat{j}-\frac{2}{\sqrt{40+(x+2{)}^{2}}}\hat{k} \end{gathered}$

$\begin{gathered} \vec{a}\cdot \vec{d}=1 \ \frac{x+2+6-2}{\sqrt{40+(x+2{)}^{2}}}=1 \ x+6=\sqrt{40+(x+2{)}^{2}} \end{gathered}$

$\begin{gathered} (x+6{)}^{2}=40+(x+2{)}^{2} \ {x}^{2}+36+12x=40+{x}^{2}+4+4x \ 8x=8 \ \Rightarrow x=1 \ (\vec{a}	imes \vec{b})\cdot \vec{c}=[\vec{a}\vec{b}\vec{c}] \ =[[|\begin{array}{ccc}1 \quad  1 \quad  1 \ 2 \quad  4 \quad  -5 \ 1 \quad  2 \quad  3\end{array}]]| \ =1(12+10)-1(6+5)+1(4-4) \ =22-11=11 \end{gathered}$

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