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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).

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Vishal Baghel

Contributor-Level 10

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 $\frac{1}{2} | A \vec{B} * A \vec{C} |$

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Class 12th Continuity and Differentiability

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

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alok kumar singh

Contributor-Level 10

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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a year ago

Class 12th Vector Algebra

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

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Vishal Baghel

Contributor-Level 10

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

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a year ago

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

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Vishal Baghel

Contributor-Level 10

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

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

From (1) and (4), we have

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

Hence, proved.

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New answer posted

a year ago

Class 12th Continuity and Differentiability

3. Examine the following functions for continuity.

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alok kumar singh

Contributor-Level 10

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},

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

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) Given, f(x) = $\frac{x^{2} - 25}{x + 5}, x \neq - 5$

For any a $?$ { 5}

$\underset{x \to a}{l i m} f (x) = \underset{x \to a}{l i m}$ $\frac{x^{2} - 25}{x + 5} = \frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5}$ = a 5

And f(a) = $\frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5} .$

= a 5

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

So, f is continuous in its domain.

(d) Given f (a) = $| x - 5 | = {\begin{matrix} x - 5 & , if x - 5 > 0 \Rightarrow x \geq 5 \\ - (x - 5) & if x - 5 < 0 \Rightarrow x < 5 . \end{matrix}$

For x = c < 5.

f (c) = (c 5) = 5 c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = (c 5) = 5 c.

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

So f is continuous.

For x = c > 5.

f (c) = (x 5) = c 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = c

...more

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},

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

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) Given, f(x) = $\frac{x^{2} - 25}{x + 5}, x \neq - 5$

For any a $?$ { 5}

$\underset{x \to a}{l i m} f (x) = \underset{x \to a}{l i m}$ $\frac{x^{2} - 25}{x + 5} = \frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5}$ = a 5

And f(a) = $\frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5} .$

= a 5

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

So, f is continuous in its domain.

(d) Given f (a) = $| x - 5 | = {\begin{matrix} x - 5 & , if x - 5 > 0 \Rightarrow x \geq 5 \\ - (x - 5) & if x - 5 < 0 \Rightarrow x < 5 . \end{matrix}$

For x = c < 5.

f (c) = (c 5) = 5 c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = (c 5) = 5 c.

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

So f is continuous.

For x = c > 5.

f (c) = (x 5) = c 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = c 5.

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

So, f is continuous.

For x = c = 5,

f (5) = 5 5 = 0

$\underset{x \to 5^{}}{l i m}$ f(a) = $\underset{x \to 5^{}}{l i m}$ (x 5) = (5 5) = 0

$\underset{x \to 5^{+}}{l i m}$ f(x) = $\underset{x \to 5^{+}}{l i m}$ (x 5) = 5 5 = 0

∴ $\underset{x \to 5^{+}}{}$ + (x) = $\underset{x \to c^{+}}{s i n}$ + (x) = f (c)

Hence f is continuous.

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New answer posted

a year ago

Class 12th Vector Algebra

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

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Vishal Baghel

Contributor-Level 10

Given,

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

For,

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

For,

$\vec{a} * \vec{b} = 0$ , 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 can conclude that

$| \vec{a} | = 0$ or $| \vec{b} | = 0$

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New answer posted

a year ago

Class 12th Vector Algebra

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

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Vishal Baghel

Contributor-Level 10

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

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

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

Class 12th Vector Algebra

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

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Vishal Baghel

Contributor-Level 10

Show that

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

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New answer posted

a year ago

Class 12th Vector Algebra

44. If a unit vector $\vec{a}$ makes an angle π/3 with $\hat{i}, \frac{π}{4} w i t h \hat{j}$ and an acute angle θ with $\hat{k}$ then find θ and hence, the components of $\vec{a}$ .

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Vishal Baghel

Contributor-Level 10

Let $\vec{a} = (a_{1}, a_{2}, a_{3})$ as component

We know,

$\vec{a}$ is a unit vector, $| \vec{a} | = 1$

Given that,

$\vec{a}$ marks angles $\frac{π}{3}$ with $\hat{i}$ , $\frac{π}{4}$ with $\hat{j}$ and $θ$ with $\hat{k}$ acute angle.

Now,

$c o s \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \Rightarrow \frac{1}{2} = a_{1} [| \vec{a} | = 1] c o s \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \Rightarrow1/\sqrt2$
$\begin{array}{l} \frac{}{} = a_{2} \\ c o s θ = \frac{a_{3}}{| \vec{a} |} \Rightarrow a_{3} = c o s θ \end{array}$

We know,

$| \vec{a} | = 1$

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New answer posted

a year ago

Class 12th Vector Algebra

43. Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b} a n d \vec{a} - \vec{b}, w h e r e \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} a n d \vec{b} = \hat{i} - 2 \hat{j} - 2 \hat{k}$

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Vishal Baghel

Contributor-Level 10

Given,

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

A vector which is perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is given by

Say

Therefore, the unit vector is

$\begin{array}{l} \frac{\vec{c}}{| \vec{c} |} = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{16}{24} \hat{i} \pm \frac{16}{24} \hat{j} \pm \frac{8}{24} \hat{k} \\ = \pm \frac{2}{3} \hat{i} \pm \frac{2}{3} \hat{j} \pm \frac{1}{3} k \end{array}$

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