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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.

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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.

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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.

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Yes, the exam authority will open the application correction facilty after the last date of registration. Candidates who fill the incorrect information while filling up the form will be able to edit the incorrect details through the official website. According to last year, the exam authority will divided the details to be corrected in two categories, below given category 1 details can be corrected online by the candidates:

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.

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

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

(i) The position vector of point R dividing the join of P and Q. internally in

the ratio 2:1 is,

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

(ii) The position vector of the point k dividing the join of P and Q. externally in the ratio 2:1

A15. (ii)

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

Here,

Let,  $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

Then,

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