Let three vectors $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ be such that $$\vec{c}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$, $$\vec{a} \cdot \vec{c} = 7$$ and $$\vec{b}$$ is perpendicular to $$\vec{c}$$, where $$\vec{a} = -\hat{i} + \hat{j} + \hat{k}$$ and $$\vec{b} = 2\hat{i} + \hat{k}$$, then the value of $$2|\vec{a} + \vec{b} + \vec{c}|^2$$ is ______

Let

$$\vec a=-\hat i+\hat j+\hat k$$ and $$\vec b=2\hat i+\hat k$$

Also,

$$\vec c$$ is coplanar with $$\vec a$$ and $$\vec b,$$ with$$\vec a\cdot\vec c=7$$ and $$\vec b\perp\vec c.$$

Find the value of $$2|\vec a+\vec b+\vec c|^2$$

Solution:

Since

$$\vec c$$is coplanar with $$\vec a$$ and $$\vec b,$$

let

$$\vec c=x\vec a+y\vec b$$

Now compute the required dot products.

$$\vec a\cdot\vec a=(-1)^2+1^2+1^2=3$$

$$\vec b\cdot\vec b=2^2+0^2+1^2=5$$

$$\vec a\cdot\vec b=(-1)(2)+(1)(0)+(1)(1)=-1$$

Given,

$$\vec b\perp\vec c$$

Therefore,

$$\vec b\cdot\vec c=0$$

$$\vec b\cdot(x\vec a+y\vec b)=0$$

$$x(\vec a\cdot\vec b)+y(\vec b\cdot\vec b)=0$$

$$-x+5y=0$$

$$x=5y

\quad\cdots(1)$$

Also,

$$\vec a\cdot\vec c=7$$

$$\vec a\cdot(x\vec a+y\vec b)=7$$

$$x(\vec a\cdot\vec a)+y(\vec a\cdot\vec b)=7$$

$$3x-y=7$$

Substituting

$$x=5y,$$

$$15y-y=7$$

$$14y=7$$

$$y=\frac12$$

Hence,

$$x=\frac52$$

Therefore,

$$\vec c=\frac52\vec a+\frac12\vec b$$

Now,

$$\vec a+\vec b+\vec c =\vec a+\vec b+\frac52\vec a+\frac12\vec b$$

$$=\frac72\vec a+\frac32\vec b$$

$$=\frac12(7\vec a+3\vec b)$$

Hence,

$$|\vec a+\vec b+\vec c|^2=\frac14|7\vec a+3\vec b|^2$$

Using

$$|\vec u+\vec v|^2=|\vec u|^2+|\vec v|^2+2\vec u\cdot\vec v,$$

we get

$$|7\vec a+3\vec b|^2=49|\vec a|^2+9|\vec b|^2+42(\vec a\cdot\vec b)$$

$$=49(3)+9(5)+42(-1)$$

$$=147+45-42$$

$$=150$$

Therefore,

$$|\vec a+\vec b+\vec c|^2=\frac14(150)$$

$$=\frac{75}{2}$$

Hence,

$$2|\vec a+\vec b+\vec c|^2=2\times\frac{75}{2}$$

$$=75$$

Therefore, the required value is

$$\boxed{75}$$