Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

52. Let the vectors $\vec{a}$ and $\vec{b}$ such that | $\vec{a}$ | = 3 and | $\vec{b}$ | = √2/3 then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is:

(A) $\frac{π}{6}$

(B) $\frac{π}{4}$

(C) $\frac{π}{3}$

(D) $\frac{π}{2}$

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(B) Given,

$∴$ Hence, $\vec{a} \times \vec{b}$ is a unit vector if angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{4}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

51. Find the area of the parallelogram whose adjacent sides are determined by the vectors

$a = \vec{i} - \vec{j} + 3 \vec{k} a n d b = 2 \vec{i} - 7 \vec{j} + \vec{k}$

17 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given,

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

The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $| \vec{a} \times \vec{b} |$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

50. Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

17 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given,

$A (1, 1, 2), B (2, 3, 5) C (1, 5, 5)$

We have,

$\begin{array}{l} A \vec{B} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ A \vec{C} = 4 \hat{j} + 3 \hat{k} \end{array}$

The area of given triangle is $\frac{1}{2} | A \vec{B} \times A \vec{C} |$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Continuity and Differentiability

5. Prove that the function $f (x) = x^{n}$ continuous at $x = n,$ where n is a positive integer.

60 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

4. Given, f (x) = x n > n = positive.

At x = 2,

(x) = n.

$\underset{x \to n}{l i m}$ f (x) = $\underset{x \to n}{l i m}$ x n = n

∴ $\underset{x \to n}{l i m}$ f (x) = f (x)

So f is continuous at x = n.

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

49. If either $\vec{a}$ = 0 and $\vec{b}$ = 0 then $\vec{a} \times \vec{b} = 0$ Is the converse true? Justify your answer with an example.

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

We take any parallel non- zero vectors so that $\vec{a} \times \vec{b} = 0$ .

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

48. Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

16 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given,

$\begin{array}{l} \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \\ \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ ∴ (\vec{b} + \vec{c}) = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k} \\ N o w, \end{array}$

$\begin{array}{l} = \hat{i} {a_{2} (b_{2} + c_{3}) - a_{3} (b_{2} + c_{2})} - \hat{j} {a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})} \\ + \hat{k} {a_{1} (b_{2} + c_{2}) - a_{2} (b_{1} + c_{1})} \\ = \hat{i} {a_{2} b_{2} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}} - \hat{j} {a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}} \\ + \hat{k} {a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{2}} - - - - (1) \end{array}$

$\begin{array}{l} = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) - - - - (2) \\ A n d, \end{array}$

$\hat{i} (a_{2} c_{3} - a_{3} c_{2}) - \hat{j} (a_{1} c_{3} - a_{3} c_{1}) + \hat{k} (a_{1} c_{2} - a_{2} c_{1}) - - - - (3)$

Adding (2) and (3), we get

$\begin{array}{l} (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) \\ + \hat{i} (a_{2} c_{3} - a_{3} c_{2}) - \hat{j} (a_{1} c_{3} - a_{3} c_{1}) + \hat{k} (a_{1} c_{2} - a_{2} c_{1}) \\ (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i} (a_{2} b_{3} - a_{3} b_{2} + a_{2} c_{3} - a_{3} c_{2}) + \hat{j} (- a_{1} b_{3} + a_{3} b_{1} - a_{1} c_{3} + a_{3} c_{1}) \\ + \hat{k} (a_{1} b_{2} - a_{2} b_{1} + a_{1} c_{2} - a_{2} c_{1}) \\ = \hat{i} (a_{2} b_{3} + a_{2} c_{3} - a_{3} c_{2} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \\ + \hat{k} (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) - - - - - (4) \end{array}$

From (1) and (4), we have

$\vec{a} (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

Hence, proved.

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New answer posted•a year ago

Ncert Solutions Maths class 12th Continuity and Differentiability

3. Examine the following functions for continuity.

31 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

3. (a) Given, f (x) = x 5.

The given f x n is a polynernial f xn and as every pohyouraial f xn is continuous in its domain R we conclude that f (x) is continuous.

(b). Given, f(x) = $\frac{1}{x - 5}, x \neq 5$

For any a $= 3 {(5 x)}^{3 c o s 2 x} [\frac{c o s 2 x}{x} - 2 s i n 2 x l o g 5 x]$ {5},

$= \frac{x^{2} \times 1 \times}{x - 3} \frac{d}{d x} (x - 3) + l o g (x - 3) \cdot 2 x$ $\frac{1}{(x - 5)} = \frac{1}{a - 5} .$

and f(a) = $\frac{1}{a - 5}$

i e, $f (x) = - (x - 1) + [- (x - 2)] = - x + 1 - x + 2 = 3 - 2 x .$ f(x) = f(a).

Hence f is continuous in its domain.

(c)

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

47. Given that $\vec{a} . \vec{b} = 0$ and $\vec{a} \times \vec{b} = 0$ What can you conclude about the vectors $\vec{a}$ and $\vec{b}$ ?

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given,

$\vec{a} . \vec{b} = 0$ and $\vec{a} \times \vec{b} = 0$

For,

$\vec{a} . \vec{b} = 0$ , then either $| \vec{a} | = 0$ or $| \vec{b} | = 0$ or $\vec{a} ⊥ \vec{b}$

For,

$\vec{a} \times \vec{b} = 0$ , then either $| \vec{a} | = 0$ or $| \vec{b} | = 0$ or $\vec{a} ? \vec{b}$

$∴$ In case $\vec{a}$ and $\vec{b}$ are non- zero on both side.

But $\vec{a}$ and $\vec{b}$ cannot be both perpendicular and parallel simultaneously.

So, we c

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

46. Find λ and μ if $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0}$

$\begin{array}{l} \Rightarrow \hat{i} (6 μ - 27 λ) - \hat{j} (2 μ - 27) + \hat{k} (2 λ - 6) \\ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \end{array}$

On comparing both side components,

$\begin{array}{l} 6 μ - 27 λ = 0, \\ 2 μ - 27 = 0 \\ 2 μ = 27 \\ μ = \frac{27}{2}, \\ 2 λ - 6 = 0 \\ 2 λ = 6 \\ λ = \frac{6}{2} = 3 \end{array}$

$∴$ Therefore, the value of $μ = \frac{27}{2}$ and $λ = 3$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Vector Algebra

45. Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

10 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Show that

$\begin{array}{l} (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b}) \\ (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \\ = \vec{a} (\vec{a} + \vec{b}) - \vec{b} (\vec{a} + \vec{b}) \\ = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \\ = 0 + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - 0 \\ = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} [\vec{a} \times \vec{b} = - \vec{b} \times \vec{a}] \\ = 2 (\vec{a} \times \vec{b}) \end{array}$

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Ncert Solutions Maths class 12th

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