Let Z, a= i+j-k and b=3 i-j+2 k. Let c be a vector such that (a+b… - Vidyadip

MCQ

Let $\lambda \in Z, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. Let $\overrightarrow{ c }$ be a vector such that $(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$ and $\vec{b} \cdot \vec{c}=-20$ Then $|\overrightarrow{ c } \times(\lambda \hat{i}+\hat{j}+\hat{ k })|^2$ is equal to

A
$62$
✓
$46$
C
$53$
D
$49$

✓

Answer

Correct option: B.

$46$

b
$(\vec{a}+\vec{b}+\overrightarrow{ c }) \times \overrightarrow{ c }=0$

$(\vec{a}+\vec{b}) \times \vec{c}=0$

$\overrightarrow{ c }=\alpha(\overrightarrow{ a }+\overrightarrow{ b })=\alpha(\lambda+3) \hat{ i }+\alpha \hat{ k }$

$\vec{b} \cdot \vec{c}=-20 \Rightarrow 3 \alpha(\lambda+3)+2 \alpha=-20$

$\vec{a} . \vec{c}=-17 \Rightarrow \alpha \lambda(\lambda+3)-\alpha=-17$

$\Rightarrow \alpha(3 \lambda+9+2)=-20$

$\alpha\left(\lambda^2+3 \lambda-1\right)=-17$

$17(3 \lambda+11)=20\left(\lambda^2+3 \lambda-1\right)$

$20 \lambda^2+9 \lambda-207=0$

$\lambda=3 \quad(\lambda \in Z)$

$\Rightarrow \alpha=-1 \quad \Rightarrow \overrightarrow{ c }=-(6 \hat{ i }+\hat{ k })$

$\overrightarrow{ v }=\overrightarrow{ c } \times(3 \hat{ i }+\hat{ j }+\hat{ k })$

$=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1\end{array}\right|=\hat{i}+3 \hat{j}-6 \hat{k}$

$|\overrightarrow{ v }|^2=(-1)^2+3^2+6^2=46$

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