Let a=i+2j+k,b=3i-3j+3k,c=2i-j+2k and d be a vector such that b d=c d and a d=4.

Question

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k},\vec{b}=3\hat{i}-3\hat{j}+3\hat{k},\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{d}$ be a vector such that $\vec{b}	imes \vec{d}=\vec{c}	imes \vec{d}$ and $\vec{a}\cdot \vec{d}=4$. Then $|(\vec{a}	imes \vec{d}){|}^{2}$ is equal to\_\_\_\_\_\_\_\_\_\_\_.

Answer: 128

Accepted Answer

Given Conditions :

The condition $\vec{b}	imes \vec{d}=\vec{c}	imes \vec{d}$ implies that $(\vec{b}-\vec{c})	imes \vec{d}=0$.

This indicates that $\vec{d}$ is parallel to $\vec{b}-\vec{c}$.

Expressing $\vec{b}-\vec{c}$:

Calculate $\vec{b}-\vec{c}=(3\hat{i}-3\hat{j}+3\hat{k})-(2\hat{i}-\hat{j}+2\hat{k})=\hat{i}-2\hat{j}+\hat{k}$.

Express $\vec{d}$:

Since $\vec{d}$ is parallel to $\vec{b}-\vec{c}$, let $\vec{d}=\lambda (\hat{i}-2\hat{j}+\hat{k})$.

Using the condition $\vec{a}\cdot \vec{d}=4$:

Compute $\vec{a}\cdot \vec{d}=(\hat{i}+2\hat{j}+\hat{k})\cdot (\lambda (\hat{i}-2\hat{j}+\hat{k}))$.

This simplifies to $\lambda (1\cdot 1+2\cdot (-2)+1\cdot 1)=\lambda (1-4+1)=-2\lambda $.

Set $-2\lambda =4$, which gives $\lambda =-2$.

Determine $\vec{d}$ using $\lambda $:

Therefore, $\vec{d}=-2(\hat{i}-2\hat{j}+\hat{k})=-2\hat{i}+4\hat{j}-2\hat{k}$.

Calculate $\vec{a}	imes \vec{d}$:

Use the determinant form for the cross product:

$\vec{a}	imes \vec{d}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  1 \ -2 \quad  4 \quad  -2 \end{vmatrix}$

Calculate the determinant:

For $\hat{i}$: $(2	imes -2)-(1	imes 4)=-4-4=-8$

For $\hat{j}$: $(-(1	imes -2)-(1	imes -2))=2-2=0$

For $\hat{k}$: $(1	imes 4)-(2	imes -2)=4+4=8$

Thus, $\vec{a}	imes \vec{d}=-8\hat{i}+0\hat{j}+8\hat{k}$.

Calculate the magnitude squared:

Find $|\vec{a}	imes \vec{d}{|}^{2}=(-8{)}^{2}+{0}^{2}+{8}^{2}=64+0+64=128$.

Therefore, $|(\vec{a}	imes \vec{d}){|}^{2}=128$.

Solutions · 1