Question

Let $\vec a$=$\hat i+\hat j+2\hat k$, $\vec b$=3$\hat i$-2$\hat j$+7$\hat k$, and $\vec c$=2$\hat i$-$\hat j$+4$\hat k$.Find a vector which is perpendicular to both $\vec a$and $\vec b$,and $\vec c$.$\vec d$=15.

Answer 1

Let d→=d1$\hat i$+d2$\hat j$+d3$\hat k$.
Since,$\vec d$ is perpendicular to both $\vec a$ and $\vec b$, we have:
$\vec d.\vec a=0$
⇒d1+4d2+7d3=0...(i)
And,
$\vec d.\vec b$=0
⇒3d1-2d2+7d3=0...(ii)
Also,it is given that:
$\vec c.\vec d$=15
⇒2d1-d2+4d3=0=15...(iii)
On solving (i),(ii),and (iii), we get:
d1=$\frac{160}{3}$,d2=$-\frac{5}{3}$,and d3=$-\frac{70}{3}$
∴$\vec d$=$\frac{160}{3} \hat i -\frac{5}{3}\hat j-\frac{70}{3}\hat k$=$\frac{1}{3}$(160$\hat i$-5$\hat j$-70$\hat k$)
Hence, the required vector is $\frac{1}{3}$(160$\hat i$-5$\hat j$-70$\hat k$).