Vector Algebra

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

Let $\theta$ be the angle between the vectors $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ .

It is given that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|,\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}and\theta =60^{?}----(1)$

We know, $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$\begin{array}{l}\therefore \frac{1}{2}=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|cos60^{?}(using(1))\\ \frac{1}{2}={\left|\overrightarrow{a}\right|}^2*\frac{1}{2}\\ {\left|\overrightarrow{a}\right|}^2=1\\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1\end{array}$

$\therefore$ Magnitude of two vector=1

$\left(3\overrightarrow{a}-5\overrightarrow{b}\right).\left(2\overrightarrow{a}+7\overrightarrow{b}\right).$

$\begin{array}{l}=3\overrightarrow{a}.2\overrightarrow{a}+3\overrightarrow{a}.7\overrightarrow{b}-5\overrightarrow{b}.2\overrightarrow{a}-5\overrightarrow{b}.7\overrightarrow{b}\\ =6\overrightarrow{a}.\overrightarrow{a}+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35\overrightarrow{b}.\overrightarrow{b}\\ =6{\left|\overrightarrow{a}\right|}^2+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\\ =6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\end{array}$

$\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ ,if $\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8$ and $\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|$

$\begin{array}{l}\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ and\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|\\ \left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ \overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}=8\\ {\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ {\left(8\left|\overrightarrow{b}\right|\right)}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 64{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 63{\left|\overrightarrow{b}\right|}^2=8\\ \left|\overrightarrow{b}\right|=\surd 8/\surd 63\left(\therefore magnitudeofavectorisnon-negative\right)\\ \left|\overrightarrow{b}\right|=\frac{\mathrm{2\surd 2}}{\mathrm{3\surd 7}}\\ And\\ \left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|=\frac{2*\mathrm{2\surd 2}}{\mathrm{3\surd 7}}=\frac{1\mathrm{6\surd 2}}{\mathrm{3\surd 7}}\end{array}$

Here, each of the given three vector is a unit vector.

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=\frac{2}{7}*\frac{3}{7}+\frac{3}{7}*\left(\frac{-6}{7}\right)+\frac{6}{7}*\frac{2}{7}\\ =\frac{6}{49}+\left(\frac{-18}{49}\right)+\frac{12}{49}=\frac{6-18+12}{49}=0\\ \overrightarrow{b}.\overrightarrow{c}=\frac{3}{7}*\frac{6}{7}+\left(\frac{-6}{7}\right)*\frac{2}{7}+\frac{2}{7}*\left(-\frac{3}{7}\right)\\ =\frac{18}{49}-\frac{12}{49}+\left(\frac{-6}{49}\right)=\frac{18-12-6}{49}=0\\ \overrightarrow{c}.\overrightarrow{a}=\frac{6}{7}*\frac{2}{7}+\frac{2}{7}*\frac{3}{7}+\left(-\frac{3}{7}\right)*\frac{6}{7}\\ =\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0\end{array}$

Therefore, the given three vectors are mutually perpendicular to each other.

Let,

$\begin{array}{l}\overrightarrow{a}=\widehat{i}+3\widehat{j}+7\widehat{k}\\ \overrightarrow{b}=7\widehat{i}-\widehat{j}+8\widehat{k}\end{array}$

The project of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is.

Let,

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

The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by,

$\therefore$ The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is 0.

Given,

=√3

$,|\overrightarrow{b}$
$=2and\overrightarrow{a}.\overrightarrow{b}=\surd 6$

We have,

We know,

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two collinear vector, they are parallel.

So,

$\begin{array}{l}\overrightarrow{b}=\lambda \overrightarrow{a}\\ If,\lambda =\pm 1,then,\overrightarrow{a}=\pm \overrightarrow{b}\\ If,\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\\ \overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k},then\\ \overrightarrow{b}=\lambda \overrightarrow{a}\\ b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=\lambda \left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\right)\\ =\left(\lambda a_1\right)\hat{i}+\left(\lambda a_2\right)\hat{j}+\left(\lambda a_3\right)\hat{k}\\ b_1=\lambda a_1,b_2=\lambda a_2,b_3=\lambda a_3\\ \frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}=\lambda \end{array}$

Hence, the respective component are proportional but, vector $\overrightarrow{a}$ and $\overrightarrow{b}$ can have different direction.

Thus, the statement given in D is incorrect.

The correct answer is D.