Class 12th

Kindly go through the solution

Vector $\lambda \overrightarrow{a}$ is a unit vector if $\left|\lambda \overrightarrow{a}\right|=1$

Now,

$\begin{array}{l}\left|\lambda \overrightarrow{a}\right|=1\\ \left|\lambda \right|\left|\overrightarrow{a}\right|=1\\ \left|\overrightarrow{a}\right|=\frac{1}{\left|\lambda \right|}\left[\lambda \ne 0\right]\\ a=\frac{1}{\left|\lambda \right|}\left[\left|\overrightarrow{a}\right|=a\right]\end{array}$

$\therefore$ Therefore, vectar $\lambda \overrightarrow{a}$ is a unit vector if a= $\frac{1}{\left|\lambda \right|}$ .

Option (D)is correct.

Let vector $2\widehat{i}-\widehat{j}+\widehat{k},\widehat{i}-3\widehat{j}-5\widehat{k}$ and $3\widehat{i}-4\widehat{j}-4\widehat{k}$ be position vector of point A, B, C respectively.

So,

$\begin{array}{l}O\overrightarrow{A}=2\widehat{i}-\widehat{j}+\widehat{k}\\ O\overrightarrow{B}=\widehat{i}-3\widehat{j}-5\widehat{k}\\ O\overrightarrow{C}=3\widehat{i}-4\widehat{j}-4\widehat{k}\end{array}$

Now, vectors $A\overrightarrow{B},B\overrightarrow{C}$ and $A\overrightarrow{C}$ represents the sides of $?ABC$ .

Hence,

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.