Maths

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.

46. The given system of inequality is

3x+4y ≤ 60 - (1)

x+3y ≤ 30- (2)

x≥ 0 - (3)

xy≥ 0 - (4)

The corresponding equation of (1) and (2) are

3x + 4y = 60

and x + 3y = 30

Putting (x, y)= (0,0) in equality (1) and (2),

3 * 0+4 * 0 ≤ 60 and 0+3 * 0 ≤ 30

0 ≤ 60 which is true and 0 ≤ 30 which is true

So, the solution plane of inequality (1) and (2) is the plane including origin (0,0)

∴ The shaded portion is the solution of the given system of inequality.

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.

(A) $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}$

By triangle law of addition in given triangle, we get:

$\begin{array}{l}\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}-----(1)\\ \overrightarrow{AB}+\overrightarrow{BC}=-\overrightarrow{CA}\end{array}$

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}-----(2)$

So, (A) is true.

(B) $\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=0$

$\begin{array}{l}\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\\ \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=0\end{array}$

So, (B) is true.

(C) $\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=0$

$\begin{array}{l}\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{CA}-----(3)\\ From,(1)\&(3),\\ \overrightarrow{AC}=\overrightarrow{CA}\\ \overrightarrow{AC}=-\overrightarrow{AC}\\ \overrightarrow{AC}+\overrightarrow{AC}=0\\ 2\overrightarrow{AC}=0\end{array}$

$\therefore$ The eQ.uation in alternative C $\overrightarrow{AC}=\overrightarrow{0}$ , which is not true, is incorrect.

(D) $\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=\overrightarrow{0}$

$\begin{array}{l}From,eqn(2)wehave\\ \overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=0\end{array}$

The, equation given is alternative is D is true.

$\therefore$ The correct answer is C.

We have,

$\begin{array}{l}\overrightarrow{a}=3\hat{i}-4\hat{j}-4\hat{k}\\ \overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}\\ \overrightarrow{c}=\hat{i}-3\hat{j}-5\hat{k}\end{array}$

$\begin{array}{l}\overrightarrow{AB}=(2-3)\hat{i}+(-1-(-4))\hat{j}+(1-(-4))\hat{k}\\ =-\hat{i}+3\hat{j}+5\hat{k}\\ \overrightarrow{BC}=(1-2)\hat{i}+(-3-(-1))\hat{j}+(-5-1)\hat{k}\\ =-\hat{i}-2\hat{j}-6\hat{k}\\ \overrightarrow{CA}=(1-3)\hat{i}+(-3-(-4))\hat{j}+(-5-(-4))\hat{k}\\ =-2\hat{i}+\hat{j}-\hat{k}\end{array}$

Now,$\begin{array}{l}\\ \end{array}$

Hence,

${\left|\overrightarrow{AB}\right|}^2+{\left|\overrightarrow{CA}\right|}^2=35+6=41={\left|\overrightarrow{BC}\right|}^2$

Hence, given points from the vertices of a right angled triangle.

The Position vector of mid-point R of the vector joining point P (2,3,4) and Q (4,1, -2) is given by;

$\begin{array}{l}\overrightarrow{OR}=\frac{\left(2\hat{i}+3\hat{j}+4\hat{k}\right)+\left(4\hat{i}+\hat{j}-\widehat{2k}\right)}{2}\\ =\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}\\ =\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=\frac{6\hat{i}}{2}+\frac{4\hat{j}}{2}+\frac{2\hat{k}}{2}\\ =3\hat{i}+2\hat{j}+\stackrel{}{k}\end{array}$