⇒Vectors

Correct answer option is (B)

Here, $\overline{c}=2\hat{i}-\hat{j}+4\hat{k}$

And given that $\overline{c}\cdot \overline{d}=15$

We verify given options one by one to satisfy the above condition.

Consider option (B),

For $\bar{\mathrm{d}}=90\hat{\mathrm{i}}-3\hat{\mathrm{j}}-42\hat{\mathrm{k}}$

$\begin{array}{rl}\overline{c}\cdot \overline{d}& =(2)(90)+(-1)(-3)+(4)(-42)\\ & =180+3-168=15\end{array}$

$\therefore$ Option (B) is correct.

Correct answer option is (D)

$\begin{array}{ll}& \bar{\mathrm{a}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}},\bar{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\\ \therefore & |\bar{\mathrm{a}}|=\sqrt{4+1+4}=3\\ & \bar{\mathrm{a}}\times \bar{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|=2\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\\ \therefore & |\bar{\mathrm{a}}\times \bar{\mathrm{b}}|=\sqrt{4+4+1}=3\end{array}$

Angle between $\bar{\mathrm{c}}$ and $\bar{\mathrm{a}}\times \bar{\mathrm{b}}$ is $\frac{\pi }{6}$ .... [Given]

$\begin{array}{rl}\therefore & \sin \frac{\pi }{6}=\frac{|(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\times \bar{\mathrm{c}}|}{|\bar{\mathrm{a}}\times \bar{\mathrm{b}}||\bar{\mathrm{c}}|}\\ & \frac{1}{2}=\frac{3}{3\times |\bar{\mathrm{c}}|}\Rightarrow |\bar{\mathrm{c}}|=2\end{array}$

Now, $|\bar{\mathrm{c}}-\bar{\mathrm{a}}|=4$ ..... [Given]

$\begin{array}{rl}& \Rightarrow |\overline{c}{|}^2+|\overline{a}{|}^2-2\overline{a}\cdot \overline{c}=16\\ & \Rightarrow 4+9-2a\cdot c=16\\ & \Rightarrow a\cdot c=\frac{-3}{2}\end{array}$

Correct answer option is (D)

According to the given condition, we get

$\begin{array}{ll}& (\bar{\mathrm{a}}+\lambda \bar{\mathrm{b}})\cdot \bar{\mathrm{c}}=0\\ \therefore & [(2+2\lambda )\hat{\mathrm{i}}+(3+\lambda )\hat{\mathrm{j}}+(2-\lambda )\hat{\mathrm{k}}]\cdot (3\hat{\mathrm{i}}-\hat{\mathrm{j}})=0\\ \therefore & 3(2+2\lambda )-(3+\lambda )=0\\ \therefore & 6+6\lambda -3-\lambda =0\\ \therefore & 3+5\lambda =0\\ \therefore & \lambda =\frac{-3}{5}\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& |\overline{a}|=1,|\overline{b}|=8,|\overline{c}|=4\text{, and }\\ & \bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}+\bar{\mathrm{c}})+\bar{\mathrm{b}}\cdot (\bar{\mathrm{c}}+\bar{\mathrm{a}})+\bar{\mathrm{c}}\cdot (\bar{\mathrm{a}}+\bar{\mathrm{b}})=0\\ & \Rightarrow 2(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{a}})=0\text{.... (i)}\end{array}$

Now,

$\begin{array}{rl}& |\overline{a}+\overline{b}+\overline{c}{|}^2=|\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2+2(\overline{a}\cdot \overline{b}+\overline{b}\cdot \overline{c}+\overline{c}\cdot \overline{a})\\ & \Rightarrow |\overline{a}+\overline{b}+\overline{c}{|}^2=1+64+16+0\dots [\text{ From (i) ] }\\ & \Rightarrow |\overline{a}+\overline{b}+\overline{c}{|}^2=81\\ & \Rightarrow |\overline{a}+\overline{b}+\overline{c}|=9\end{array}$

Correct answer option is (B)

Since $\overline{a}+2\overline{b}$ and $5\overline{a}-4\overline{b}$ are perpendicular to each other

$\begin{array}{rl}\therefore & (\bar{\mathrm{a}}+2\bar{\mathrm{b}})\cdot (5\bar{\mathrm{a}}-4\bar{\mathrm{b}})=0\\ & \Rightarrow 5|\bar{\mathrm{a}}{|}^2-8|\bar{\mathrm{b}}{|}^2+6\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}=0\\ \Rightarrow & -3+6|\bar{\mathrm{a}}||\bar{\mathrm{b}}|\cos \theta =0\dots [\because |\bar{\mathrm{a}}|=|\bar{\mathrm{b}}|=1]\\ \Rightarrow & \cos \theta =\frac{1}{2}\\ \Rightarrow & \theta =\frac{\pi }{3}\end{array}$