Let Z,a= i+j-k and b=3i-j+2k.

Question

Let $\lambda \in Z,\vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}+\vec{c})	imes \vec{c}=\overset{	o }{0},\vec{a}\cdot \vec{c}=-17$ and $\vec{b}\cdot \vec{c}=-20$. Then $|\vec{c}	imes (\lambda \hat{i}+\hat{j}+\hat{k}){|}^{2}$ is equal to :

Accepted Answer

The given vectors are :

$\vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$

$\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$

We are given that $(\vec{a}+\vec{b}+\vec{c})	imes \vec{c}=0$ which implies $(\vec{a}+\vec{b})	imes \vec{c}=0$. So, $\vec{c}$ is in the direction of $\vec{a}+\vec{b}$.

Let's denote $\vec{c}=\alpha (\vec{a}+\vec{b})$.

Substituting values for $\vec{a}$ and $\vec{b}$, we get :

$\vec{c}=\alpha ((\lambda +3)\hat{i}+\hat{k})$

From the conditions $\vec{a}\cdot \vec{c}=-17$ and $\vec{b}\cdot \vec{c}=-20$, we can form two equations :

$\alpha \lambda (\lambda +3)-\alpha =-17\, \, \alpha (3\lambda +9+2)=-20.$

By solving the above equations, we find $\alpha =-1$ and $\lambda =3$.

Substituting these values back into $\vec{c}$ gives $\vec{c}=-6\hat{i}-\hat{k}$.

Next, we need to find $|\vec{c}	imes (\lambda \hat{i}+\hat{j}+\hat{k}){|}^{2}$.

This simplifies to :

$|\vec{c}	imes (3\hat{i}+\hat{j}+\hat{k}){|}^{2}$

Calculate the cross product $\vec{c}	imes (3\hat{i}+\hat{j}+\hat{k})$ which gives :

$=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ -6 \quad  0 \quad  -1 \ 3 \quad  1 \quad  1 \end{vmatrix}=\hat{i}+3\hat{j}-6\hat{k}$

Finally, the square of the magnitude of this vector is:

$|\vec{c}	imes (3\hat{i}+\hat{j}+\hat{k}){|}^{2}=(1{)}^{2}+{3}^{2}+(-6{)}^{2}=1+9+36=46.$

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