⇒Vectors

Correct answer option is (C)

$\begin{array}{rl}& |\bar{\mathrm{a}}\times \bar{\mathrm{b}}{|}^2+(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}{)}^2=144\text{ and }|\bar{\mathrm{a}}|=4\\ & \therefore (|\mathrm{a}{|}^2\cdot |\mathrm{b}{|}^2\cdot {\sin }^2\theta )+(|\bar{\mathrm{a}}{|}^2|\bar{\mathrm{b}}{|}^2{\cos }^2\theta )=144\\ & \therefore |\bar{\mathrm{a}}{|}^2|\bar{\mathrm{b}}{|}^2({\cos }^2\theta +{\sin }^2\theta )=144\\ & \therefore (4{)}^2|\bar{\mathrm{b}}{|}^2=144\Rightarrow |\bar{\mathrm{b}}{|}^2=9\Rightarrow |\bar{\mathrm{b}}|=3\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& \overline{c}=\overline{a}\times \overline{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -5& 0\\ 6& 3& 0\end{array}\right|=\hat{i}(0)-\hat{j}(0)+\hat{k}(9+30)=39\hat{k}\\ & |\overline{a}|=\sqrt{(3{)}^2+(-5{)}^2}=\sqrt{34}\text{ and }|\overline{b}|=\sqrt{(6{)}^2+(3{)}^2}=\sqrt{45}\text{ and }|\overline{c}|=\sqrt{{39}^2}=39\\ & \therefore a:b:c=\sqrt{34}:\sqrt{45}:39\end{array}$

Correct answer option is (D)

We have $|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|=20$ and we have to find value of $|(3\bar{\mathrm{a}}+\bar{\mathrm{b}})\times (2\bar{\mathrm{a}}+3\bar{\mathrm{b}})|$

$\begin{array}{rl}& (3\bar{\mathrm{a}}+\bar{\mathrm{b}})\times (2\bar{\mathrm{a}}+3\bar{\mathrm{b}})\\ & =6(\bar{\mathrm{a}}\times \bar{\mathrm{a}})+9(\bar{\mathrm{a}}\times \bar{\mathrm{b}})+2(\bar{\mathrm{b}}\times \bar{\mathrm{a}})+3(\bar{\mathrm{b}}\times \bar{\mathrm{b}})\\ & =0+9(\bar{\mathrm{a}}\times \bar{\mathrm{b}})-2(\bar{\mathrm{a}}\times \bar{\mathrm{b}})+0=7(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\end{array}$

Hence area of the required parallelogram $=7\times 20=140$

Correct answer option is (C)

From given data, we write

$\begin{array}{rl}& \left|\begin{array}{ccc}\hat{i}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 6& 27\\ 1& \lambda & \mu \end{array}\right|=0\\ & \therefore \hat{\mathrm{i}}(6\mu -27\lambda )-\hat{\mathrm{j}}(2\mu -27)+\hat{\mathrm{k}}(2\lambda -6)=0\end{array}$

$\therefore$ 6$\mu$ $-$ 27$\lambda$ = 0 ..... (1)

27 $-$ 2$\mu$ = 0 ..... (2)

2$\lambda$ $-$ 6 = 0 ..... (3)

From (2) and (3), we get $\mu =\frac{27}{2}$ and $\lambda =3$.

These values of $\lambda$ & $\mu$ satisfy eq. (1)