Ncert Solutions Maths class 12th

Minimize $z=3x+5y$

Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+3y=3\\ x+y=2\\ x,y=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{3}+\frac{y}{1}=1\\ \frac{x}{2}+\frac{y}{2}=1\\ x,y=0\end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A\left(3,0\right),B\left(\frac{3}{2},\frac{1}{2}\right)\&C\left(0,2\right)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum value of Z.

We draw the graph of inequality $3x+5y<7$ .

The feasible region has no common point with $3x+5y<7$ .

Therefore minimum value of Z in 7 at $B\left(\frac{3}{2},\frac{1}{2}\right)$

Maximise $z=5x+3y$

Subject to $2x+5y\le 15,5x+2y\le 10,x\ge 0,y\ge 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l}3x+5y=15\\ 5x+2y=10\&x=0,y=0\end{array}$

$\Rightarrow \frac{x}{5}+\frac{y}{3}=1$

$\begin{array}{l}\frac{x}{2}+\frac{y}{5}=1\\ x=0,y=0\end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O\left(0,0\right),A\left(2,0\right),B\left(\frac{20}{19},\frac{45}{19}\right)\&C\left(0,3\right)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$

Minimize $z=-3x+4y$

Subject to $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=8\\ 3x+2y=12\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$

$\begin{array}{l}\frac{x}{4}+\frac{y}{6}=1\\ x=0,y=0\end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O (0,0), A (4,0), B (2,3), and C (0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0).

Maximise $z=3x+4y$

Subject to the constraints: $x+y\le 4,x\ge 0,y\ge 0$

The corresponding equation of the above inequality are

$\begin{array}{l}x+y=4\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{4}+\frac{y}{4}=1$

$x=0,y=0$

The graph of the given inequalities.

The shaded region OAB is the feasible region which is bounded.

The corresponding of the corner point of the feasible region are O (0,0), A (4,0), and B (0,4).

The value of Z at these points are as follows,

Corner point $z=3x+4y$

O (0,0) 0

A (4,0) 12

B (0,4) 16

Therefore the maximum value of Z is 16 at the point B (0,4).

$\hat{i}\left(\hat{j}*\hat{k}\right)+\hat{j}\left(\hat{i}*\hat{k}\right)+\hat{k}\left(\hat{i}*\hat{j}\right)$

$\begin{array}{l}=\hat{i}.\hat{i}+\hat{j}\left(-\hat{j}\right)+\hat{k}.\hat{k}\\ =1-\hat{j}.\hat{j}+1\\ =1-1+1\\ =1\end{array}$

Therefore, the correct answer is (C)

Let, $\overrightarrow{a}\&\overrightarrow{b}$ be two unit vectors and $\theta$ be the angle between them.

Then, $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1$

Now, $\overrightarrow{a}+\overrightarrow{b}$ is a unit vector if $\left|\overrightarrow{a}+\overrightarrow{b}\right|=1$

$\begin{array}{l}\left|\overrightarrow{a}+\overrightarrow{b}\right|=1\\ {(\overrightarrow{a}+\overrightarrow{b})}^2=1\\ (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=1\\ \overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}=1\\ {\left|\overrightarrow{a}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2=1\\ 1^2+2\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\theta +1^2=1\end{array}$

$1+2.1.1cos\theta +1=1$ [ $\therefore$ $\overrightarrow{a}\&\overrightarrow{b}$ is unit vector.]

$\begin{array}{l}2cos\theta =1-2\\ cos\theta =\frac{-1}{2}=\frac{2\pi }{3}\\ \therefore \theta =\frac{2\pi }{3}\end{array}$

Therefore, the correct answer is (D)

Let, $\theta$ in triangle between two vector $\overrightarrow{a}\&\overrightarrow{b}$

Then, without loss of generality, $\overrightarrow{a}\&\overrightarrow{b}$ are non-zero vector so that $\left|\overrightarrow{a}\right|\&\left|\overrightarrow{b}\right|$

are positive.

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

So, $\overrightarrow{a}.\overrightarrow{b}\ge 0$

$\begin{array}{l}\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta \ge 0\\ cos\theta \ge 0\left[\therefore \left|\overrightarrow{a}\right|\&\left|\overrightarrow{b}\right|arepositive\right]\\ 0\le \theta \le \frac{\pi }{2}\end{array}$

Therefore, $\overrightarrow{a}.\overrightarrow{b}\ge 0$ , when $0\le \theta \le \frac{\pi }{2}$

Hence, the correct answer is B.

$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2$

$\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b}={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2$

( Distributive of scalar product over addition )

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

( Scalar product is commutative , $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}$ )

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

$\therefore$ Therefore, $\overrightarrow{a}\&\overrightarrow{b}$ are perpendicular.

Given that $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{c}$ are mutually perpendicular vectors, we have

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0\\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|\end{array}$

Let, vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ be inclined to $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{c}$ at angles, ${\theta }_1,{\theta }_2\&{\theta }_3$ respectively.

$\begin{array}{l}\therefore cos{\theta }_1=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{a}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{a}\right|}=\frac{\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{a}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{a}\right|}\\ =\frac{{\left|\overrightarrow{a}\right|}^2}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{a}\right|}\left[\overrightarrow{b}.\overrightarrow{a}=\overrightarrow{c}.\overrightarrow{a}=0\right]\\ =\frac{\left|\overrightarrow{a}\right|}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}\end{array}$

$\begin{array}{l}\therefore cos{\theta }_2=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{b}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{b}\right|}\\ =\frac{\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{b}\right|}\left[\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=0\right]\\ =\frac{{\left|\overrightarrow{b}\right|}^2}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{b}\right|}=\frac{\left|\overrightarrow{b}\right|}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}\\ \therefore cos{\theta }_3=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{c}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{c}\right|}\\ =\frac{\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{c}}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{c}\right|}\left[\overrightarrow{a}.\overrightarrow{c}=\overrightarrow{b}.\overrightarrow{c}=0\right]\\ =\frac{{\left|\overrightarrow{c}\right|}^2}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|\left|\overrightarrow{c}\right|}=\frac{\left|\overrightarrow{c}\right|}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}\\ now,as,\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|,\\ cos{\theta }_1=cos{\theta }_2=cos{\theta }_3\\ \therefore {\theta }_1={\theta }_2={\theta }_3\end{array}$

Therefore, the vector $\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$ are equally inclined to $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{\mathrm{c.}}$