Question

What is the unit vector perpendicular to the following vectors 2i+2j-k and 6i-3j+2k

Answer 1

Answer: $\frac{i-10j-18k}{5\sqrt{17} }$

Explanation:

A=2i+2j-k         ;B=6i-3j+2k

To find a unit vector'N' perpendicular to a combination of two vectors 'A' and 'B' we do

N=(A×B)/(|A×B|)

A×B=$\left[\begin{array}{ccc}i&j&k\\2&2&-1\\6&-3&2\end{array}\right]$

      = i[(2×2)-(-1)×(-3)] - j[(2×2)-(6×(-1))] + k[((-3)×2)-(6×2)

      = i-10j-18k

|A×B|=$\sqrt{1^{2} +(-10)^{2} +(-18)^{2} }$

        =$\sqrt{10+100+324}$=$\sqrt{425}$

        =5$\sqrt{17}$

∴ unit vector perpendicular to A & B is

N=(A×B)/(|A×B|)=$\frac{i-10j-18k}{5\sqrt{17} }$

Answer 2

Answer:

Unit vector perpendicular to the given vectors is $\frac{\hat{i}-10 \hat{j}-18\hat{k} }{5\sqrt{17} }$

Explanation:

Given vectors $\vec A=2\hat{i}+2\hat{j}-\hat{k}$ and $\vec A=6\hat{i}-3\hat{j}+2\hat{k}$

$\vec A\times \vec B\ =\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&2&-1\\6&-3&2\end{array}\right|$

        $= \big[(2\times2)-(-1\times-3)\big ]\[ \hat{i}-\big[(2\times2)-(6\times-1)\big ]\[ \hat{j}+\big[(-3\times2)-(6\times2)\big ]\hat{k}$

        $= \big[4-3\big ]\[ \hat{i}-\big[4-(-6)\big ]\[ \hat{j}+\big[(-6)-(12)\big ]\hat{k}= \hat{i}-10 \hat{j}-18\hat{k}$

Unit vector perpendicular to the vector $\vec A\times \vec B$ is

                  $\vec A\times \vec B = \frac{\vec A\times \vec B }{\big|\vec A\times \vec B \big| }$

                             $= \frac{\hat{i}-10 \hat{j}-18\hat{k} }{\sqrt{\hat{i}-10 \hat{j}-18\hat{k}} }$

                            $= \frac{\hat{i}-10 \hat{j}-18\hat{k} }{\sqrt{1^{2}+(-10)^{2} +(-18)^{2}} }= \frac{\hat{i}-10 \hat{j}-18\hat{k} }{\sqrt{1+100 +324} }$

                            $= \frac{\hat{i}-10 \hat{j}-18\hat{k} }{5\sqrt{17} }$

Unit vector perpendicular to the vectors A and B is $\frac{\hat{i}-10 \hat{j}-18\hat{k} }{5\sqrt{17} }$