Vector Algebra Question 4
Question 4 - 2024 (27 Jan Shift 1)
Let $\vec{a}=\hat{i}+2 \hat{j}+k, \vec{b}=3(\hat{i}-\hat{j}+k)$. Let $\vec{c}$ be the vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$. Then $\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})$ is equal to :
(1) 32
(2) 24
(3) 20
(4) 36
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Answer (2)
Solution
$\vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}]$
$\vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}$
given $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$
$\Rightarrow(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{b}}|^{2}=27$
$\Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=\left[\begin{array}{lll}\vec{a} & \vec{c} & \vec{b}\end{array}\right]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \ldots$ (ii)
Now $\vec{a} \cdot \vec{b}=3-6+3=0$
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$…(iv)(given)
By (i), (ii), (iii) $\backslash &$ (iv)
$27-0-3=24$
