Class 12th

Given, point are

Vertices of $?ABC$ are given as

$\begin{array}{l}A\left(1,2,3\right),B\left(-1,0,0\right),C\left(0,1,2\right)\\ \end{array}$

$\therefore ?ABC$ is the angle between the vectors $B\overrightarrow{A}$ and $B\overrightarrow{C}$

Consider

$\begin{array}{l}\overrightarrow{a}=2\widehat{i}+4\widehat{j}+3\widehat{k}\\ \overrightarrow{b}=3\widehat{i}+3\widehat{j}-6\widehat{k}\end{array}$ and

Then,

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=2.3+4.3+3.(-6)\\ =6+12-18=0\end{array}$

Therefore, the converse of the given statement need not be true.

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right).\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$

$\begin{array}{l}=\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}\\ ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ =1+1+1+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ =3+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ \Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=\frac{-3}{2}\end{array}$

2. Given, f (x) = 2x2 1

At x = 3

Lim f (x) = $\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$ 2 (3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

We know,

$\overrightarrow{a}.\overrightarrow{a}=0$ and $\overrightarrow{a}.\overrightarrow{b}=0$

Now,

$\overrightarrow{a}.\overrightarrow{a}=0\Rightarrow {\left|\overrightarrow{a}\right|}^2\Rightarrow \left|\overrightarrow{a}\right|=0$

$\therefore$ $\overrightarrow{a}$ is a zero vector.

Thus, vector $\overrightarrow{b}$ satisfying $\overrightarrow{a}.\overrightarrow{b}=0$ can be any vector.

$\left(\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}\right).\left(\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}\right)$

$\begin{array}{l}=\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{b}\right|\overrightarrow{a}+\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{b}\right|\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2\overrightarrow{b}.\overrightarrow{b}-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}.\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2{\left|\overrightarrow{a}\right|}^2\\ =0\end{array}$

$\therefore$ Therefore, $\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}$ and $\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}$ are perpendicular.

Given,

$\begin{array}{l}\overrightarrow{a}=2\widehat{i}+2\widehat{j}+3\widehat{k}\\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}\\ \overrightarrow{c}=3\widehat{i}+\widehat{j}\end{array}$

Now,

$\begin{array}{l}\overrightarrow{a}+\lambda \overrightarrow{b}=\left(2\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(-\widehat{i}+2\widehat{j}+\widehat{k}\right)\\ =\left(2\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\left(-\lambda \widehat{i}+2\lambda \widehat{j}+\lambda \widehat{k}\right)\\ =\left(2-\lambda \right)\widehat{i}+\left(2+2\lambda \right)\widehat{j}+\left(3+\lambda \right)\widehat{k}\end{array}$

If $\left(\overrightarrow{a}+\lambda \overrightarrow{b}\right)$ is perpendicular to $\overrightarrow{c}$ , then $\left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\overrightarrow{c}=0$

$\begin{array}{l}=\left[\left(2-\lambda \right)\widehat{i}+\left(2+2\lambda \right)\widehat{j}+\left(3+\lambda \right)\widehat{k}\right].\left(3\widehat{i}+\widehat{j}\right)\\ =3\left(2-\lambda \right)+1\left(2+2\lambda \right)+0\left(3+\lambda \right)\\ =6-3\lambda +2+2\lambda +0\\ =8-\lambda \\ \Rightarrow \lambda =8\end{array}$

Therefore, the required value of $\lambda$ is 8.

$\begin{array}{l}\left(\overrightarrow{x}-\overrightarrow{a}\right).\left(\overrightarrow{x}-\overrightarrow{a}\right)=12\\ \overrightarrow{x}.\overrightarrow{x}+\overrightarrow{x}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{x}-\overrightarrow{a}.\overrightarrow{a}=12\\ {\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=12\\ {\left|\overrightarrow{x}\right|}^2-1=12\left[\left|\overrightarrow{a}\right|=1asa\overrightarrow{a}isunitvector\right]\\ {\left|\overrightarrow{x}\right|}^2=12+1=13\\ \therefore \left|\overrightarrow{x}\right|=\surd 13\end{array}$