Question
$\text{Find}\ \lambda\ \text{and}\ \mu\ \text{if}\ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}.$
$6\mu-27\lambda=0\ \ \ \ \ .....\text{(i)}$
$2\mu-27=0\ \ \ \ \ \ .....\text{(ii)}$
$\text{And}\ \ 2\lambda-6=0\ \ \ \ \ \ .....\text{(iii)}$ $\text{From eq. (ii),}\ \ 2\mu-27=0 \ \Rightarrow\ \ \mu=\frac{27}{2}$ $\text{From eq. (iii),}\ \ 2\lambda-6=0\ \Rightarrow\ \ \lambda=\frac{6}{2}=3$ 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$ $\text{Therefore,}\ \ \mu=\frac{27}{2}\ \text{and}\ \lambda=\frac{6}{2}=3$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{f(x)}=\sin^{-1}\text{x}+\sin^{-1}2\text{x}$