Let x be a vector in the plane containing vectors a=2 i- j+ k and b= i+2 j- k. If the vector x is perpendicular to (3 i+2 j- k) and its projection on…

A plane containing two vectors can be expressed as a linear combination of the vectors.

Hence, let $\vec{x}=\lambda \vec{a}+\mu \vec{b}(\lambda$x=λa+μb(λ and $\mu$μ are scalars$).$).

$$\Rightarrow \vec{x}=\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu )$$⇒x=i^(2λ+μ)+j^​(2μ−λ)+k^(λ−μ)

Since $\vec{x}$x is perpendicular to $3\hat{i}+2\hat{j}-\hat{k},$3i^+2j^​−k^,

$$\Rightarrow \vec{x}\cdot (3\hat{i}+2\hat{j}-\hat{k})=0$$⇒x⋅(3i^+2j^​−k^)=0

$$\Rightarrow (\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu ))\cdot (3\hat{i}+2\hat{j}-\hat{k})=0$$⇒(i^(2λ+μ)+j^​(2μ−λ)+k^(λ−μ))⋅(3i^+2j^​−k^)=0

$$\Rightarrow 3(2\lambda +\mu )+2(2\mu -\lambda )-(\lambda -\mu )=0$$⇒3(2λ+μ)+2(2μ−λ)−(λ−μ)=0

$$\Rightarrow 3\lambda +8\mu =0...(1)$$⇒3λ+8μ=0...(1)

Also, the projection of $\vec{x}$x on $\vec{a}$a is $\frac{17\sqrt{6}}{2}$2176​​

$$\Rightarrow \frac{\vec{x}\cdot \vec{a}}{|\vec{a}|}=\frac{17\sqrt{6}}{2}$$⇒∣a∣x⋅a​=2176​​

$$\Rightarrow \frac{(\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu ))\cdot (2\hat{i}-\hat{j}+\hat{k})}{\sqrt{{2}^{2}+{1}^{2}+{1}^{2}}}=\frac{17\sqrt{6}}{2}$$⇒22+12+12​(i^(2λ+μ)+j^​(2μ−λ)+k^(λ−μ))⋅(2i^−j^​+k^)​=2176​​

$$\Rightarrow \frac{2(2\lambda +\mu )-(2\mu -\lambda )+(\lambda -\mu )}{\sqrt{6}}=\frac{17\sqrt{6}}{2}$$⇒6​2(2λ+μ)−(2μ−λ)+(λ−μ)​=2176​​

$$\Rightarrow 6\lambda -\mu =51...(2)$$⇒6λ−μ=51...(2)

On solving the equations $(1)$(1) and $(2),$(2), we get $\lambda =8,\mu =-3.$λ=8,μ=−3.

Thus, $\vec{x}=13\hat{i}-14\hat{j}+11\hat{k}$x=13i^−14j^​+11k^

$$\Rightarrow |\vec{x}|=\sqrt{{13}^{2}+{(-14)}^{2}+{11}^{2}}$$⇒∣x∣=132+(−14)2+112​

$$\Rightarrow |\vec{x}{|}^{2}=486.$$⇒∣x∣2=486.