Let a=i+j+k,b=3i+2j-k,c= j+ k and d be a unit vector such that a d=b d and c d=1.

Question

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

Answer: 5

Accepted Answer

Given the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$, along with $\vec{c}=\lambda \hat{j}+\mu \hat{k}$, and $\hat{d}$ being a unit vector such that $\vec{a}	imes \hat{d}=\vec{b}	imes \hat{d}$ and $\vec{c}\cdot \hat{d}=1$, we proceed as follows:

Determine $\hat{d}$:

Since $\vec{a}	imes \hat{d}=\vec{b}	imes \hat{d}$, we have:

$(\vec{a}-\vec{b})	imes \hat{d}=0$

Thus, $\hat{d}$ is parallel to $\vec{a}-\vec{b}$. Calculate:

$\vec{a}-\vec{b}=(-2\hat{i}+\hat{j}+2\hat{k})$

Therefore, we can express $\hat{d}$ as:

$\hat{d}=\frac{-2}{3}\hat{i}-\frac{1}{3}\hat{j}+\frac{2}{3}\hat{k}$

Solve for $\lambda $ and $\mu $:

Using the condition $\vec{c}\cdot \hat{d}=1$:

$\frac{-\lambda }{3}+\frac{2\mu }{3}=1$

Simplify to:

$-\lambda +2\mu =3\,$

Since $\vec{c}$ is perpendicular to $\vec{a}$:

$\vec{c}\cdot \vec{a}=0$

Which gives:

$\lambda +\mu =0\, \Rightarrow \lambda =-\mu $

Substitute $\lambda =-\mu $ into equation (i):

$\mu +2\mu =3\, \Rightarrow 3\mu =3\, \Rightarrow \mu =1$

Therefore, $\lambda =-1$.

Find $|3\lambda \hat{d}+\mu \vec{c}{|}^{2}$:

Calculate:

$|3\lambda \hat{d}+\mu \vec{c}{|}^{2}=9{\lambda }^{2}|\hat{d}{|}^{2}+{\mu }^{2}|\vec{c}{|}^{2}+2\cdot 3\cdot \lambda \cdot \mu \cdot \vec{c}\cdot \vec{d}$

Substituting the determined values:

$|\hat{d}{|}^{2}=1$ (since $\hat{d}$ is a unit vector)

$\vec{c}\cdot \vec{d}=1$

Thus:

$=9(-1{)}^{2}\cdot 1+{1}^{2}\cdot ({1}^{2}+{1}^{2})+2\cdot 3\cdot (-1)\cdot 1\cdot 1$

$=9+2-6$

$=5$

Hence, $|3\lambda \hat{d}+\mu \vec{c}{|}^{2}=5$.

Solutions · 1