Once upon a time on a plane, there lived three vectors...

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There are thee vectors $\vec{a}, \vec{b}, \vec{c}$

Given that $\vec{a} = p\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + q\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} + r\hat{k}$

Given that these three vectors are coplanar and $p \neq q \neq r \neq 1$

Then let the value of $pqr - ( p+q+r)$ be $k - 10$

Find K

1 solution

Deepansh Mathur
Mar 29, 2014

The given vectors are collinear if

$\begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} = 0$

$\Rightarrow p(qr-1) + 1(1-r) + 1(1-q) = 0$

$\Rightarrow pqr - p + 1 - r + 1 - q = 0$

$\Rightarrow pqr - (p + q + r) = -2$

So $-2 = k - 10$ , which gives $k = 8$