If a=i+j+k,b=j-k and c be three vectors such that a c=b and a c=3, then c (a-2b) is equal …

Question

If $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a}	imes \vec{c}=\vec{b}$ and $\vec{a}\cdot \vec{c}=3$, then $\vec{c}\cdot (\vec{a}-2\vec{b})$ is equal to $____$ .

Answer: 3

Accepted Answer

Let

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

We have to find

$\vec{c}\cdot (\vec{a}-2\vec{b})$

given that

$\vec{a}	imes \vec{c}=\vec{b}\, \, \vec{a}\cdot \vec{c}=3$

Let

$\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$

Then

$\vec{a}=(1,1,1),\, \vec{b}=(0,1,-1)$

Now compute $\vec{a}	imes \vec{c}$:

$\vec{a}	imes \vec{c}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  1 \quad  1 \ x \quad  y \quad  z \end{vmatrix}$

Expanding,

$\vec{a}	imes \vec{c}=\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)$

Since this is equal to $\vec{b}=\hat{j}-\hat{k}$, we compare components:

$z-y=0$

$-(z-x)=1\Rightarrow x-z=1$

$y-x=-1$

Also, from $\vec{a}\cdot \vec{c}=3$,

$x+y+z=3$

Now solve the equations.

From $z-y=0$,

$z=y$

From $x-z=1$,

$x=y+1$

Substitute into $x+y+z=3$:

$(y+1)+y+y=3$

$3y+1=3$

$3y=2\Rightarrow y=\frac{2}{3}$

So,

$z=\frac{2}{3},\, x=\frac{5}{3}$

Hence,

$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$

Now find $\vec{a}-2\vec{b}$:

$\vec{a}-2\vec{b}=(\hat{i}+\hat{j}+\hat{k})-2(\hat{j}-\hat{k})$

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

Therefore,

$\vec{c}\cdot (\vec{a}-2\vec{b})=(\frac{5}{3},\frac{2}{3},\frac{2}{3})\cdot (1,-1,3)$

$=\frac{5}{3}-\frac{2}{3}+2$

$=\frac{3}{3}+2=1+2=3$

So, the required value is

$3$

Solutions · 1