Let a=i+4j+2k,b=3i-2j+7k and c=2i-j+4k.

Question

Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k},\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$. If a vector $\vec{d}$ satisfies $\vec{d}	imes \vec{b}=\vec{c}	imes \vec{b}$ and $\vec{d}\cdot \vec{a}=24$, then $|\vec{d}{|}^{2}$ is equal to :

Accepted Answer

Given that $\vec{d}	imes \vec{b}=\vec{c}	imes \vec{b}$, we can rewrite this as:

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

This implies that the vector $\vec{d}-\vec{c}$ is a scalar multiple of $\vec{b}$:

$\vec{d}=\vec{c}+\lambda \vec{b}$

Also, we are given that $\vec{d}\cdot \vec{a}=24$:

$(\vec{c}+\lambda \vec{b})\cdot \vec{a}=24$

Now, we can find the value of $\lambda $ :

$\lambda =\frac{24-\vec{a}\cdot \vec{c}}{\vec{b}\cdot \vec{a}}=\frac{24-6}{9}=2$

Therefore, we have :

$\vec{d}=\vec{c}+2(\vec{b})=8\hat{i}-5\hat{j}+18\hat{k}$

Now, we can find the squared magnitude of $\vec{d}$ :

$|\vec{d}{|}^{2}=64+25+324=413$

Solutions · 1