Question

If $\vec a= 2\hat i +\hat j + 3\hat k$, $\vec b = 3\hat i + 3\hat j + \hat k$ and $\vec c = c_1\hat i + c_2\hat j + c_3\hat k$ are coplanar vectors and $\vec a.\vec c = 5$, $\vec b⊥\vec c$ then $122( c_1 + c_2 + c_3 )$ is equal to _____ .

Answer 1

$2C_1 + C_2 \+ 3C_3 = 5$ …(i)
$3C_1 \+ 3C_2 + C_3 = 0$ …(ii)
$[ \vec a \vec b\vec c] = \begin{vmatrix} 2 & 1 & 3 \\\[0.3em] 3 & 3 & 1 \\\[0.3em] C_1 & C_2 & C_3 \end{vmatrix}$
$= 2(3C_3 – C_2) – 1(3C_3 – C_1) + 3(3C_2 – 3C_1)$
$= 3C_3 \+ 7C_2 – 8C_1$
$⇒ 8C_1 – 7C_2 – 3C_3 = 0$ …(iii)
$C_1=$ $\frac {10}{122}$,

C_2 =$$-\frac {85}{122}

$C_3 =$ $\frac {225}{122}$

$122(C_1 + C_2 + C_3)$
= $122$ ($\frac {10}{122}$ $-\frac {85}{122}$ +$\frac {225}{122}$ )
= $10-85+225$
= $150$

So, the answer is $150$.