Let a = i + j + k , b =3 i +2 j - k , c = j + k and d be a unit v… - Vidyadip

Question

Let $\overrightarrow{ a }=\hat{ i }+\hat{ j }+\hat{ k }, \overrightarrow{ b }=3 \hat{ i }+2 \hat{ j }-\hat{ k }, \overrightarrow{ c }=\lambda \hat{ j }+\mu \hat{ k }$ and $\hat{ d }$ be a unit vector such that $\overrightarrow{ a } \times \hat{ d }=\overrightarrow{ b } \times \hat{ d }$ and $\overrightarrow{ c } . \hat{ d }=1$, If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \overrightarrow{ c }|^2$ is equal to __________.

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Answer

(5)
$\begin{array}{l}\overrightarrow{ a } \times \overrightarrow{ d }-\overrightarrow{ b } \times \overrightarrow{ d }=0 \\ ( a -\overrightarrow{ b }) \times \overrightarrow{ d }=0 \\ \overrightarrow{d}= t (\overrightarrow{ a }-\overrightarrow{ b }) \\ \overrightarrow{ d }= t (-2 \hat{ i }-\hat{ j }+2 \hat{ k }) \\ |\overrightarrow{ d }|=1 \\ | t |=\frac{1}{3}\end{array}$
$\begin{array}{l}\overrightarrow{ c } \cdot \overrightarrow{ a }=0 \\ \lambda+\mu=0 \\ \mu=-\lambda \\ \overrightarrow{ c }=\lambda(\hat{ j }-\hat{ k }),|\overrightarrow{ c }|^2=2 \lambda^2 \\ \overrightarrow{ c } \cdot \hat{ d }=1 \\ t (-2,-1,2) \cdot \lambda(0,1,-1)=1 \\ \lambda t =\frac{-1}{3} \Rightarrow \lambda^2=1\end{array}$
$\begin{array}{l}|3 \lambda \hat{d}+\mu c |^2=9 \lambda^2|\hat{d}|^2+\mu^2|\overrightarrow{ c }|^2+6 \lambda \mu(\hat{d} \cdot \overrightarrow{ c }) \\ =3 \lambda^2+2 \lambda^4 \\ =5\end{array}$

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