⇒Vectors

Correct answer option is (A)

From given data, we write

$\overline{a}+\lambda \overline{b}=(2-\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+\lambda )\hat{k}$ ..... (1)

Since (1) is $\perp$ er to $\overline{c}$, we write

$\begin{array}{rl}& (2-\lambda )(3)+(2+2\lambda )(1)+(3+\lambda )(2)=0\\ & \therefore 6-3\lambda +2+2\lambda +6+2\lambda =0\Rightarrow \lambda =-14\end{array}$

Correct answer option is (C)

$|\bar{\mathrm{a}}+\bar{\mathrm{b}}{|}^2=|\bar{\mathrm{a}}-\bar{\mathrm{b}}{|}^2+4\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}$ ..... (1)

Now $|\bar{\mathrm{a}}-\bar{\mathrm{b}}{|}^2=|\bar{\mathrm{a}}{|}^2+|\mathrm{b}{|}^2-2\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}$

$\therefore (5{)}^2=(3{)}^2+(4{)}^2-2\overline{a}\cdot \overline{b}\Rightarrow \overline{a}\cdot \overline{b}=0$

Substituting in (1), we get

$|\bar{\mathrm{a}}+\bar{\mathrm{b}}{|}^2=|\bar{\mathrm{a}}-\bar{\mathrm{b}}{|}^2\Rightarrow |\bar{\mathrm{a}}+\bar{\mathrm{b}}|=5$

Correct answer option is (D)

We have $\overline{a}\cdot (\bar{\mathrm{b}}+\bar{\mathrm{c}})=0,\bar{\mathrm{b}}\cdot (\bar{\mathrm{c}}+\bar{\mathrm{a}})=0$ and $\bar{\mathrm{c}}\cdot (\bar{\mathrm{a}}+\bar{\mathrm{b}})=0$

$\therefore \bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{a}}\cdot \bar{\mathrm{c}}=0$ ....... (1)

$\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{a}}=0$ ..... (2)

$\bar{\mathrm{c}}\cdot \bar{\mathrm{a}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{b}}=0$ ..... (3)

From (1), (2) and (3), we get

$\begin{array}{rl}& 2(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{a}})=0\\ & \text{ Now }|\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}{|}^2=|\bar{\mathrm{a}}{|}^2+|\bar{\mathrm{b}}{|}^2+|\bar{\mathrm{c}}{|}^2+2(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{a}})\\ & \therefore |\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}{|}^2=(5{)}^2+(4{)}^2+(3{)}^2+2(0)=50\end{array}$

Correct answer option is (C)

$\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}=0\Rightarrow \bar{\mathrm{c}}=-(\bar{\mathrm{a}}+\bar{\mathrm{b}})$ and let angle between $\bar{\mathrm{a}}$ and $\bar{\mathrm{b}}$ be $\theta$

$\begin{array}{rl}\therefore |\overline{c}{|}^2& =(\overline{a}+\overline{b}{)}^2=|\bar{\mathrm{a}}{|}^2+|\bar{\mathrm{b}}{|}^2+2\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}\\ & =|\bar{\mathrm{a}}{|}^2+|\bar{\mathrm{b}}{|}^2+2|\bar{\mathrm{a}}|\cdot |\bar{\mathrm{b}}|\cdot \cos \theta \\ \therefore (7{)}^2& =(3{)}^2+(5{)}^2+2(3)(5)\cos \theta \\ \therefore 49& =9+25+30\cos \theta \Rightarrow \cos \theta =\frac{15}{30}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}\end{array}$

Correct answer option is (B)

${|{\bar{\mathrm{e}}}_1+{\bar{\mathrm{e}}}_2|}^2=\left|{\bar{\mathrm{e}}}_1^2\right|+\left|{\bar{\mathrm{e}}}_2^2\right|+2{\bar{\mathrm{e}}}_1{\bar{\mathrm{e}}}_2\cos \theta$

Here $\left|{\bar{\mathrm{e}}}_1\right|=1,\left|{\bar{\mathrm{e}}}_2\right|=1$ and $|{\bar{\mathrm{e}}}_1+{\bar{\mathrm{e}}}_2|=1$

$\begin{array}{rl}& \therefore (1)=(1{)}^2+(1{)}^2+2(1)(1)\cos \theta \\ & \therefore \frac{-1}{2}=\cos \theta \Rightarrow \theta ={120}^{\circ }\end{array}$