If a = i + j - k , b =- i +2 j +2 k and c =- i +2 j - k , then a unit vector normal to the vectors a + b and b - c is: 1. i 2. j 3. k 4. None of these

1. i Solution: a + b =0 i +3 j + k b - c =0 i -0 j +3 k ( a b ) ( b - c )= i & j & k 0&3&1 0&0&3 =9 i | ( a + b ) ( b - c ) |=9| i | =9(1) =9 Unit vector perpendicular to both a + b and b - c = ( a + b ) ( b - c ) | ( a + b ) ( b - c ) | = 9 i 9 = i

If a = i + j - k , b =- i +2 j +2 k and c =- i +2 j - k , then a …

Download App

Question

If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ then a unit vector normal to the vectors $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}$ is:

$\hat{\text{i}}$
$\hat{\text{j}}$
$\hat{\text{k}}$
$\text{None of these}$

✓

Answer

$\hat{\text{i}}$

Solution:
$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$
$=9\hat{\text{i}}$
$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$
$=9(1)$
$=9$
Unit vector perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)}{\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|}$
$=\frac{9\hat{\text{i}}}{9}$
$=\hat{\text{i}}$