Question

Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat i-3\hat j+6 \hat k),\frac{1}{7}(3\hat i-6\hat j+2 \hat k), \frac{1}{7}(6\hat i-2\hat j-3 \hat k)$Also, show that they are mutually perpendicular to each other.

Answer 1

we know, magnitude of unit vector equals 1.

here given three vectors

$\vec{A}= \frac{1}{7}(2\hat i-3\hat j+6 \hat k)$

$\vec{B}=\frac{1}{7}(3\hat i+6\hat j+2 \hat k)$

$\vec{C}=\frac{1}{7}(6\hat i-2\hat j-3 \hat k)$

Let $\vec{A}=\frac{1}{7}(2\hat i-3\hat j+6 \hat k)=\frac{2}{7}\hat i-\frac{3}{7}\hat j+\frac{6}{7}\hat k$

magnitude of $\vec{A}$ = $|\vec{A}|$

= $\sqrt{\left(\frac{2}{7}\right)^2+\left(\frac{-3}{7}\right)^2+\left(\frac{6}{7}\right)^2}$

=$\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$

similarly, you can find, $|\vec{B}|=|\vec{C}|=1$ ,by using above method.

now, we have to prove that they are mutually perpendicular to each other.

it is possible when $\vec{A}.\vec{B}=0$ and $\vec{B}.\vec{C}=0$

$\vec{A}.\vec{B}=\frac{1}{7}(2\hat i-3\hat j+6 \hat k).\frac{1}{7}(3\hat i+6\hat j+2 \hat k)$

= $\frac{1}{49}\{2\times3+(-3)\times(6)+6\times2\}$

= $\frac{1}{49}\{6-18+12\}=0$

and $\vec{B}.\vec{C}=\frac{1}{7}(3\hat i+6\hat j+2 \hat k).\frac{1}{7}(6\hat i-2\hat j-3 \hat k)$

= $\frac{1}{49}\{3\times6+6\times-2+2\times-3\}$

=$\frac{1}{49}\{18-12-6\}=0$

hence, $\vec{A},\vec{B},\textbf{and},\vec{C}$ are mutually perpendicular.