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Vector Algebra

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New answer posted

2 months ago

Vector Algebra Class 12th

If m is the slope of a common tangent to the curves $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 a n d x^{2} + y^{2} = 12, t h e n$ $\frac{^{}}{} \frac{^{}}{}^{}^{}$ 12 m² is equal to:

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Vishal Baghel

Contributor-Level 10

Equation of any tangent to $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

is y = mx + $\sqrt[]{16 m^{2} + 9}$ if this line is also tangent to x² + y² = 12

then $\sqrt[]{12}$ = $| \frac{\sqrt[]{16 m^{2} + 9}}{\sqrt[]{1 + m^{2}}} |$

^{$\Rightarrow 12 + 12 m^{2} = 16 m^{2} + 9$}

$\Rightarrow 12 m^{2} = 9$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $Δ P Q R$ is

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Payal Gupta

Contributor-Level 10

RM = $| \frac{3 + 7 - 5}{\sqrt[]{2}} | = \frac{5}{\sqrt[]{2}}$

$l s i n 60 ° = \frac{5}{\sqrt[]{2}} \Rightarrow l = 5 \sqrt[]{\frac{2}{3}}$

$∴ A r e a o f Δ P Q R = \frac{3}{4} l^{2} =$ =25/2√3

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New answer posted

2 months ago

Vector Algebra Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Let $x_{0}$ be the point of local maxima of $f (x) = \vec{a} \cdot (\vec{b} * \vec{c})$ where $\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = - 2 \hat{i} + x \hat{j} - \hat{k}$ and $\vec{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ . Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ at $x = x_{0}$ is

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alok kumar singh

Contributor-Level 10

$f (x) = \overset{?}{a} \cdot (\overset{?}{b} * \overset{?}{c}) = |\begin{matrix} x & - 2 & 3 \\ - 2 & x & - 1 \\ 7 & - 2 & x \end{matrix}|$

$= x^{3} - 27 x + 26$

$f^{'} (x) = 3 x^{2} - 27 = 0 \Rightarrow x = \pm 3$ and $f^{''} (- 3) < 0$

$\Rightarrow$ local maxima at $x = x_{0} = - 3$

Thus, $\overset{?}{a} = - 3 \hat{i} - 2 \hat{j} + 3 \hat{k}, \overset{?}{b} = 2 \hat{i} - 3 \hat{j} - \hat{k}$ , and $\overset{?}{c} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k}$

$\Rightarrow \overset{?}{a} \cdot \overset{?}{b} + \overset{?}{b} \cdot \overset{?}{c} + \overset{?}{c} \cdot \overset{?}{a} = 9 - 5 - 26 = - 22$

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New answer posted

2 months ago

Vector Algebra Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , then the value of $| \hat{i} * (\vec{a} * \hat{i}) |^{2} + | \hat{j} * (\vec{a} * \hat{j}) |^{2} + | \hat{k} * (\vec{a} * \hat{k}) |^{2}$ is equal to

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alok kumar singh

Contributor-Level 10

Given ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{r} \frac{2^{10} {(2 x^{2} + 3 x + 4)}^{10}}{x^{20}} = \sum_{r = 0}^{20} ? ? \frac{a_{r} 2^{r}}{x^{r}} \\ \Rightarrow 2^{10} \sum_{r = 0}^{20} ? ? a_{r} x^{r} = \sum_{r = 0}^{20} ? ? a_{r} 2^{r} x^{(20 - r)} (from (i)) \end{array}$

now, comparing coefficient of $x^{7}$ from both sides

(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)

$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

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New answer posted

4 months ago

Vector Algebra 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

4 months ago

Vector Algebra 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

4 months ago

Vector Algebra 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

4 months ago

Vector Algebra 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 7 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

4 months ago

Vector Algebra 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

4 months ago

Vector Algebra 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 $\vec{c}$ .

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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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