3D Geometry 3 Question 31
31. Let $\mathbf{u}$ be a vector coplanar with the vectors $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{u}$ is perpendicular to $\mathbf{a}$ and $\mathbf{u} \cdot \mathbf{b}=24$, then $|\mathbf{u}|^{2}$ is equal to
(2018 Main)
(a) 336
(b) 315
(c) 256
(d) 84
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Answer:
Correct Answer: 31. (d)
Solution:
Key Idea If any vector $x$ is coplanar with the vector $y$ and $z$, then $\mathbf{x}=\lambda \mathbf{y}+\mu \mathbf{z}$
Here, $\mathbf{u}$ is coplanar with $\mathbf{a}$ and $\mathbf{b}$.
$ \therefore \quad \mathbf{u}=\lambda \mathbf{a}+\mu \mathbf{b} $
Dot product with a, we get
$ \mathbf{u} \cdot \mathbf{a}=\lambda(\mathbf{a} \cdot \mathbf{a})+\mu(\mathbf{b} \cdot \mathbf{a}) \Rightarrow 0=14 \lambda+2 \mu $
$ [\because \mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{u} \cdot \mathbf{a}=0] $
Dot product with $\mathbf{b}$, we get
$ \begin{aligned} & \mathbf{u} \cdot \mathbf{b}=\lambda(\mathbf{a} \cdot \mathbf{b})+\mu(\mathbf{b} \cdot \mathbf{b}) \\ & 24=2 \lambda+2 \mu \end{aligned} $
Solving Eqs. (i) and (ii), we get
$ \lambda=-2, \mu=14 $
Dot product with $\mathbf{u}$, we get
$ \begin{aligned} & |\mathbf{u}|^{2}=\lambda(\mathbf{u} \cdot \mathbf{a})+\mu(\mathbf{u} \cdot \mathbf{b}) \\ & \\ & |\mathbf{u}|^{2}=-2(0)+14(24) \Rightarrow|\mathbf{u}|^{2}=336 \end{aligned} $
