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New answer posted

a year ago

Maths Linear Programming

Solve the following Linear Programming Problems graphically:

1. Maximize Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

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Vishal Baghel

Contributor-Level 10

Maximise $z = 3 x + 4 y$

Subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0$

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 $z = 3 x + 4 y$

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).

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New answer posted

a year ago

Maths Vector Algebra

72. If θ be the angle between any two vectors $\vec{a}$ and $\vec{b}$ , then $| \vec{a} . \vec{b} | = | \vec{a} * \vec{b} |$ when θ is equal to:

(A) 0

(B) $\frac{π}{4}$

(C) $\frac{π}{2}$

(D) π

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Vishal Baghel

Contributor-Level 10

Let, $θ$ be angle between two vector $\vec{a} & \vec{b}$ .

Then, without loss of generality, $\vec{a} & \vec{b}$ are non-zero vectors, so that $\vec{a} & \vec{b}$

are positive.

$\begin{array}{l} | \vec{a} . \vec{b} | = | \vec{a} * \vec{b} | \\ | \vec{a} | | \vec{b} | c o s θ = | \vec{a} | | \vec{b} | s i n θ \\ c o s θ = s i n θ [| \vec{a} | & | \vec{b} | a r e p o s i t i v e] \\ t a n θ = 1 = \frac{π}{4} ∴ θ = \frac{π}{4} . \end{array}$

$∴$ Therefore, the correct answer is B.

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New answer posted

a year ago

Maths Vector Algebra

71. Choose the correct answer:

The value of $\hat{i} . (\hat{j} * \hat{k}) + \hat{j} . (\hat{i} * \hat{k}) + \hat{k} (\hat{i} * \hat{j})$ is:

(A) 0

(B) -1

(C) 1

(D) 3

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Vishal Baghel

Contributor-Level 10

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

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

Therefore, the correct answer is (C)

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New answer posted

a year ago

Maths Vector Algebra

70. Choose the correct answer:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors andθ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if

(A) θ = $\frac{π}{4}$

(B) θ = $\frac{π}{3}$

(C) θ = $\frac{π}{2}$

(D) θ = $\frac{π}{3}$

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Vishal Baghel

Contributor-Level 10

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

Then, $| \vec{a} | = | \vec{b} | = 1$

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

$\begin{array}{l} | \vec{a} + \vec{b} | = 1 \\ {(\vec{a} + \vec{b})}^{2} = 1 \\ (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = 1 \\ \vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{b} . \vec{a} + \vec{b} . \vec{b} = 1 \\ {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} = 1 \\ 1^{2} + 2 | \vec{a} | . | \vec{b} | c o s θ + 1^{2} = 1 \end{array}$

$1 + 2.1.1 c o s θ + 1 = 1$ [ $∴$ $\vec{a} & \vec{b}$ is unit vector.]

$\begin{array}{l} 2 c o s θ = 1 - 2 \\ c o s θ = \frac{- 1}{2} = \frac{2 π}{3} \\ ∴ θ = \frac{2 π}{3} \end{array}$

Therefore, the correct answer is (D)

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New answer posted

a year ago

Maths Vector Algebra

69. Choose the correct answer:

If θ is the angle between two vectors $\vec{a}$ and $\vec{b}$ then $\vec{a} . \vec{b} \geq 0$ only when:

(A) $0 < θ < \frac{π}{2}$

(B) $0 \leq θ \leq \frac{π}{2}$

(C) $0 < θ < π$

(D) $0 \leq θ \leq π$

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

Let, $θ$ in triangle between two vector $\vec{a} & \vec{b}$

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

are positive.

We know, $\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ$

So, $\vec{a} . \vec{b} \geq 0$

$\begin{array}{l} \Rightarrow | \vec{a} | | \vec{b} | c o s θ \geq 0 \\ c o s θ \geq 0 [∴ | \vec{a} | & | \vec{b} | a r e p o s i t i v e] \\ 0 \leq θ \leq \frac{π}{2} \end{array}$

Therefore, $\vec{a} . \vec{b} \geq 0$ , when $0 \leq θ \leq \frac{π}{2}$

Hence, the correct answer is B.

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New answer posted

a year ago

Maths Vector Algebra

68. Prove that $(\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$ if and only if $\vec{a}, \vec{b}$ are perpendicular given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$ .

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Vishal Baghel

Contributor-Level 10

$(\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

$\vec{a} . \vec{a} + \vec{a} . \vec{b} + \vec{a} . \vec{b} + \vec{b} . \vec{b} = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

( Distributive of scalar product over addition )

$\Rightarrow {| \vec{a} |}^{2} + 2 \vec{a} . \vec{b} + {| \vec{b} |}^{2} = {| \vec{a} |}^{2} + {| \vec{b} |}^{2}$

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

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

$∴$ Therefore, $\vec{a} & \vec{b}$ are perpendicular.

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New answer posted

a year ago

Maths Vector Algebra

67. If are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ -is equally inclined to $a, b,$ and $\vec{c}$ .

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0 \\ | \vec{a} | = | \vec{b} | = | \vec{c} | \end{array}$

Let, vector $\vec{a} + \vec{b} + \vec{c}$ be inclined to $\vec{a}, \vec{b} & \vec{c}$ at angles, $θ_{1}, θ_{2} & θ_{3}$ respectively.

$\begin{array}{l} ∴ c o s θ_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{a}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} = \frac{\vec{a} . \vec{a} + \vec{b} . \vec{a} + \vec{c} . \vec{a}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} \\ = \frac{{| \vec{a} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{a} |} [\vec{b} . \vec{a} = \vec{c} . \vec{a} = 0] \\ = \frac{| \vec{a} |}{| \vec{a} + \vec{b} + \vec{c} |} \end{array}$

$\begin{array}{l} ∴ c o s θ_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{b}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} \\ = \frac{\vec{a} . \vec{b} + \vec{b} . \vec{b} + \vec{b} . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} [\vec{a} . \vec{b} = \vec{b} . \vec{c} = 0] \\ = \frac{{| \vec{b} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{b} |} = \frac{| \vec{b} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ ∴ c o s θ_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} \\ = \frac{\vec{a} . \vec{c} + \vec{b} . \vec{c} + \vec{c} . \vec{c}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} [\vec{a} . \vec{c} = \vec{b} . \vec{c} = 0] \\ = \frac{{| \vec{c} |}^{2}}{| \vec{a} + \vec{b} + \vec{c} | | \vec{c} |} = \frac{| \vec{c} |}{| \vec{a} + \vec{b} + \vec{c} |} \\ n o w, a s, | \vec{a} | = | \vec{b} | = | \vec{c} |, \\ c o s θ_{1} = c o s θ_{2} = c o s θ_{3} \\ ∴ θ_{1} = θ_{2} = θ_{3} \end{array}$

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

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New answer posted

a year ago

Maths Vector Algebra

66. The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2 \hat{i} + 4 \hat{j} - 5 \hat{k} a n d λ \hat{i} + 2 \hat{j} + 3 \hat{k}$ is equal to one. Find the value of λ.

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Vishal Baghel

Contributor-Level 10

$(2 \hat{i} + 4 \hat{j} - 5 \hat{k}) + (λ \hat{i} + 2 \hat{j} + 3 \hat{k}) = (2 + λ) \hat{i} + 6 \hat{j} - 2 \hat{k}$

The unit vector along $(2 \hat{i} + 4 \hat{j} - 5 \hat{k}) + (λ \hat{i} + 2 \hat{j} + 3 \hat{k})$ is given as;

By Q.uestion, scalar product of $(\hat{i} + \hat{j} + \hat{k})$ with this unit vector is 1.

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New answer posted

a year ago

Maths Vector Algebra

65. Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} a n d \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} . \vec{d}$ = 15

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Vishal Baghel

Contributor-Level 10

Given,

$\begin{array}{l} \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k} \\ \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} \\ \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \end{array}$

Let, $\vec{d} = d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}$

Since, $\vec{d}$ is perpendicular to both $\vec{a} & \vec{b}$

$\begin{array}{l} \vec{d} . \vec{a} = 0 \\ \Rightarrow d_{1} + d_{2} 4 + d_{3} 2 = 0 - - - - - (1) \\ \vec{d} . \vec{b} = 0 \\ \Rightarrow d_{1} 3 + d_{2} (- 2) + d_{3} (7) = 0 \\ \Rightarrow d_{1} 3 - 2 d_{2} + 7 d_{3} = 0 - - - - - (2) \end{array}$

We know,

$\begin{array}{l} \vec{c} . \vec{d} = 15 \\ \Rightarrow 2 d_{1} - d_{2} + 4 d_{3} = 15 - - - - - (3) \\ F r o m, (1) \\ d_{1} + 4 d_{2} + 2 d_{3} = 0 \\ d_{1} = - 4 d_{2} - 2 d_{3} \end{array}$

Putting this value in (3) we get

$\begin{array}{l} \Rightarrow 2 (- 4 d_{2} - 2 d_{3}) - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 8 d_{2} - 4 d_{3} - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 9 d_{2} = 15 \\ \Rightarrow d_{2} = \frac{15}{9} = - \frac{5}{3} \end{array}$

Putting $d_{1} & d_{2}$ value in (2), we get

$\begin{array}{l} \Rightarrow 3 d_{1} - 2 d_{2} + 7 d_{3} = 0 \\ \Rightarrow 3 (- 4 d_{2} - 2 d_{3}) - 2 (- \frac{5}{3}) + 7 d_{3} = 0 \\ \Rightarrow - 12 * (- \frac{5}{3}) - 6 d_{3} + \frac{10}{3} + 7 d_{3} = 0 \\ \Rightarrow 20 + d_{3} + \frac{10}{3} = 0 \\ \Rightarrow d_{3} = - 20 - \frac{10}{3} = \frac{- 60 - 10}{3} = \frac{- 70}{3} \\ N o w, d_{1} = - 4 * - \frac{5}{3} - 2 * - \frac{70}{3} \\ = \frac{20}{3} + \frac{140}{3} = \frac{160}{3} \\ ∴ d_{1} = \frac{160}{3}, d_{2} = - \frac{5}{3}, d_{3} = \frac{- 70}{3} \\ \vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} \frac{- 70}{3} \hat{k} = \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k}) \end{array}$