Vectors 4 Question 1
1. Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}$ be a vector such that $\mathbf{a} \times \mathbf{c}+\mathbf{b}=\mathbf{0}$ and $\mathbf{a} \cdot \mathbf{c}=4$, then $|\mathbf{c}|^{2}$ is equal to
(2019 Main, 9 Jan I)
(a) 8
(b) $\frac{19}{2}$
(c) 9
(d) $\frac{17}{2}$
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Answer:
Correct Answer: 1. (b)
Solution:
- We have, $(a \times c)+b=0$
$\Rightarrow a \times(a \times c)+a \times b=0$
(taking cross product with a on both sides)
$$ \begin{aligned} & \Rightarrow(a \cdot c) a-(a \cdot a) c+\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{array}\right|=0 \\ & \quad[\because a \times(b \times c)=(a \cdot c) b-(a \cdot b) c] \\ & \Rightarrow 4(\hat{i}-\hat{j})-2 c+(-\hat{i}-\hat{j}+2 \hat{k})=0 \\ & \quad[\because a \cdot a=(\hat{i}-\hat{j})(\hat{i}-\hat{j})=1+1=2 \text { and } a \cdot c=4] \\ & \Rightarrow 2 c=4 i-4 \hat{j}-\hat{i}-\hat{j}+2 \hat{k} \\ & \Rightarrow c=\frac{3 \hat{i}-5 \hat{j}+2 \hat{k}}{2} \Rightarrow \quad|c|^{2}=\frac{9+25+4}{4}=\frac{19}{2} \end{aligned} $$
