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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA2 Marks
Question
$\text{Find}\ \lambda\ \text{and}\ \mu\ \text{if}\ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}.$
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Answer

$\text{Given:}\ \ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=0$ $\Rightarrow\ \ \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&6&27\\1&\lambda&\mu\end{vmatrix}=\vec{0}$ Expanding along first row, $\hat{i}(6\mu-27\lambda)-\hat{j}(2\mu-27)+\hat{k}(2\lambda-6)$ $=\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}$ Comparing the coefficients of $\hat{i}, \hat{j},\hat{k}$ on both sides, we have$6\mu-27\lambda=0\ \ \ \ \ .....\text{(i)}$
$2\mu-27=0\ \ \ \ \ \ .....\text{(ii)}$
$\text{And}\ \ 2\lambda-6=0\ \ \ \ \ \ .....\text{(iii)}$ $\text{From eq. (ii),}\ \ 2\mu-27=0 \ \Rightarrow\ \ \mu=\frac{27}{2}$ $\text{From eq. (iii),}\ \ 2\lambda-6=0\ \Rightarrow\ \ \lambda=\frac{6}{2}=3$ Putting the values of $\mu\ \text{and}\ \lambda$ in eq. (i), $6\bigg(\frac{27}{2}\bigg)-27(3)=0\ $ $ \Rightarrow\ 3(27)-27(3)=0\ \Rightarrow\ \ 0=0$ $\text{Therefore,}\ \ \mu=\frac{27}{2}\ \text{and}\ \lambda=\frac{6}{2}=3$

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