Consider the vectors overrightarrowA=i+j-k, overrightarrowB=2 i-j+k, overrightarrowC=1/(√5)(i-2 j+2 k) . What is the value of overrightarrowC .(overrightarrowA × overrightarrowB) ?

Correct answer: (B) 0. Hint: overrightarrowC(overrightarrowA × overrightarrowB)= | 1 1 -1 2 -1 1 1 -2 2 |=1(-2+2)-1(4-1)-1(-4+1)=-3+3=0

Q124 / 155 · WBJEE 2020 · Physics

Consider the vectors $\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{C}}=\frac{1}{\sqrt{5}}(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$. What is the value of $\overrightarrow{\mathrm{C}} .(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}) ?$

A
1
Correct answer
B
0
C
$3 \sqrt{2}$
D
$18 \sqrt{5}$

Solution

Hint: $\overrightarrow{\mathrm{C}}(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})=\left|\begin{array}{ccc}1 & 1 & -1 \\ 2 & -1 & 1 \\ 1 & -2 & 2\end{array}\right|=1(-2+2)-1(4-1)-1(-4+1)=-3+3=0$

Hence, the correct answer is option: 'B'.

Solution

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