Find a unit vector perpendicular to each of the vectors a + b and a - b , where a =3 i +2 j +2 k and b = i +2 j -2 k .

Given, a =3 i +2 j +2 k b = i +2 j -2 k Let, d = a + b = (3 i +2 j +2 k )+ ( i +2 j -2 k ) d =4 i +4 j -0 k And, e = a - b = (3 i +2 j +2 k )- ( i +2 j -2 k ) e =2 i +4 k Let, f be any vector perpendicular to both d and e f = i & j & k 4&4&0 2&0&4 = i (16-0)- j (16-0)+ k (0-8) f =16 i -16 j -8 k =8(2 i -2 j - k ) Let g be the required vector, then g = f and | g |=1 g =8 (2 i -2 j - k ) (1) | g |=1 8 (2)^2+(-2)^2+(-1)^2=1 8 4+4+1=1 8 9=1 24 =1 =1 24 put in (1) g =8 (1 24 ) (2 i -2 j - k ) g =1 2 (2 i -2 j - k ) Thus, Unit vector perpendicular to ( a + b ) and ( a - b )=1 3 (2 i -2 j - k )

Find a unit vector perpendicular to each of the vectors a + b and…

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Question

Find a unit vector perpendicular to each of the vectors $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}},$ where $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$

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Answer

Given, $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Let, $\vec{\text{d}}=\vec{\text{a}}+\vec{\text{b}}$
$=\big(3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)+\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{d}}=4\hat{\text{i}}+4\hat{\text{j}}-0\hat{\text{k}}$
And, $\vec{\text{e}}=\vec{\text{a}}-\vec{\text{b}}$
$=\big(3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)-\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{e}}=2\hat{\text{i}}+4\hat{\text{k}}$
Let, $\vec{\text{f}}$ be any vector perpendicular to both $\vec{\text{d}}$ and $\vec{\text{e}}$
$\vec{\text{f}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&4&0\\2&0&4 \end{vmatrix}$
$=\hat{\text{i}}(16-0)-\hat{\text{j}}(16-0)+\hat{\text{k}}(0-8)$
$\vec{\text{f}}=16\hat{\text{i}}-16\hat{\text{j}}-8\hat{\text{k}}$
$=8(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
Let $\vec{\text{g}}$ be the required vector, then
$\vec{\text{g}}=\lambda\vec{\text{f}}$ and $|\vec{\text{g}}|=1$
$\vec{\text{g}}=8\lambda\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)\dots(1)$
$|\vec{\text{g}}|=1$
$8\lambda\sqrt{(2)^2+(-2)^2+(-1)^2}=1$
$8\lambda\sqrt{4+4+1}=1$
$8\lambda\sqrt{9}=1$
$24\lambda=1$
$\lambda=\frac{1}{24}$
put $\lambda$ in (1)
$\vec{\text{g}}=8\Big(\frac{1}{24}\Big)\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)$
$\vec{\text{g}}=\frac{1}{2}\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)$
Thus,
Unit vector perpendicular to $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{a}}-\vec{\text{b}}\big)=\frac{1}{3}\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)$

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