Vector Algebra — Maths STD 12 Science — Question

$\Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&6&{27} \\ 1&\lambda &u \end{array}} \right| = \vec 0$

Expanding along first row,

$\hat i\left( {6\mu - 27\lambda } \right) - \hat j\left( {2\mu - 27} \right) + \hat k\left( {2\lambda - 6} \right) $ $= \vec 0 = 0\hat i + 0\hat j + 0\hat k$

Comparing the coefficients of $\hat i,\hat j,\hat k$ on both sides, we have

$6\mu - 27\lambda = 0$ …(i)

$2\mu - 27 = 0$ …(ii)

And $2\lambda - 6 = 0$ ...(iii)

From eq. (ii), $2\mu - 27 = 0 \Rightarrow u = \frac{{27}}{2}$

From eq. (iii), $2\lambda - 6 = 0 \Rightarrow \lambda = \frac{6}{2} = 3$

Putting the values of $\mu $ and $\lambda $ in eq. (i),

$6\left( {\frac{{27}}{2}} \right) - 27\left( 3 \right) = 0 $ $\Rightarrow 3\left( {27} \right) - 27\left( 3 \right) = 0 \Rightarrow 0 =0$

Therefore, $\mu = \frac{{27}}{2}$ and $\lambda = 3.$