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Class 12th Vector Algebra

34. Show that $| \vec{a} | \vec{b} + | \vec{b} |$ is perpendicular to $| \vec{a} | \vec{b} - | \vec{b} | \vec{a}$ for any two non-zero vectors $\vec{a}$ and $\vec{b}$

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Vishal Baghel

Contributor-Level 10

$(| \vec{a} | \vec{b} + | \vec{b} | \vec{a}) . (| \vec{a} | \vec{b} - | \vec{b} | \vec{a})$

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

$∴$ Therefore, $| \vec{a} | \vec{b} + | \vec{b} | \vec{a}$ and $| \vec{a} | \vec{b} - | \vec{b} | \vec{a}$ are perpendicular.

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New answer posted

a year ago

Class 12th Vector Algebra

33. If $\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 1 \vec{j} + \vec{k} a n d c = 3 \hat{i} + \hat{j}$ are such that $\vec{a} + λ \vec{b}$ is perpendicular to $\vec{c}$ then find the value of λ

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Vishal Baghel

Contributor-Level 10

Given,

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

Now,

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

If $(\vec{a} + λ \vec{b})$ is perpendicular to $\vec{c}$ , then $(\vec{a} + λ \vec{b}) . \vec{c} = 0$

$\begin{array}{l} = [(2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}] . (3 \hat{i} + \hat{j}) \\ = 3 (2 - λ) + 1 (2 + 2 λ) + 0 (3 + λ) \\ = 6 - 3 λ + 2 + 2 λ + 0 \\ = 8 - λ \\ \Rightarrow λ = 8 \end{array}$

Therefore, the required value of $λ$ is 8.

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New answer posted

a year ago

Class 12th Vector Algebra

32. Find $| \vec{x} |$ if for a unit vector $(\vec{x} - \vec{a}) . (\vec{x} - \vec{a}) = 12$ .

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} (\vec{x} - \vec{a}) . (\vec{x} - \vec{a}) = 12 \\ \vec{x} . \vec{x} + \vec{x} . \vec{a} - \vec{a} . \vec{x} - \vec{a} . \vec{a} = 12 \\ {| \vec{x} |}^{2} - {| \vec{a} |}^{2} = 12 \\ {| \vec{x} |}^{2} - 1 = 12 [| \vec{a} | = 1 a s a \vec{a} i s u n i t v e c t o r] \\ {| \vec{x} |}^{2} = 12 + 1 = 13 \\ ∴ | \vec{x} | =\sqrt13 \end{array}$

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New answer posted

a year ago

Class 12th Vector Algebra

31. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ having the same magnitude such that the angle between them is 60° and their scalar product is 1/2.

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Vishal Baghel

Contributor-Level 10

Let $θ$ be the angle between the vectors $| \vec{a} |$ and $| \vec{b} |$ .

It is given that $| \vec{a} | = | \vec{b} |, \vec{a} . \vec{b} = \frac{1}{2} a n d θ = 6 0^{?} - - - - (1)$

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

$\begin{array}{l} ∴ \frac{1}{2} = | \vec{a} | | \vec{a} | c o s 6 0^{?} (u s i n g (1)) \\ \frac{1}{2} = {| \vec{a} |}^{2} * \frac{1}{2} \\ {| \vec{a} |}^{2} = 1 \\ | \vec{a} | = | \vec{b} | = 1 \end{array}$

$∴$ Magnitude of two vector=1

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New answer posted

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Class 12th Continuity and Differentiability

1. Prove that the function f(x) = 5x 3f is continuous at x = 0, at x = – 3 and at x = 5.

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alok kumar singh

Contributor-Level 10

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

At x = 0, $\underset{x \to 0}{l i m} f (x) = \underset{x \to 0}{l i m}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3, $π + h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5, $\forall x \in ?$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

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New answer posted

a year ago

Class 12th Vector Algebra

30. Evaluate the product $(3 \vec{a} - 5 \vec{b}) . (2 \vec{a} + 7 \vec{b}) .$

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Vishal Baghel

Contributor-Level 10

$(3 \vec{a} - 5 \vec{b}) . (2 \vec{a} + 7 \vec{b}) .$

$\begin{array}{l} = 3 \vec{a} . 2 \vec{a} + 3 \vec{a} . 7 \vec{b} - 5 \vec{b} . 2 \vec{a} - 5 \vec{b} . 7 \vec{b} \\ = 6 \vec{a} . \vec{a} + 21 \vec{a} . \vec{b} - 10 \vec{a} . \vec{b} - 35 \vec{b} . \vec{b} \\ = 6 {| \vec{a} |}^{2} + 21 \vec{a} . \vec{b} - 10 \vec{a} . \vec{b} - 35 {| \vec{b} |}^{2} \\ = 6 {| \vec{a} |}^{2} + 11 \vec{a} . \vec{b} - 35 {| \vec{b} |}^{2} \end{array}$

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New answer posted

a year ago

Class 12th Vector Algebra

29. Find $| \vec{a} |$ and $| \vec{b} |$ ,if $(\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8$ and $| \vec{a} | = 8 | \vec{b} |$

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Vishal Baghel

Contributor-Level 10

$| \vec{a} |$ and $| \vec{b} |$ ,if $(\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8$ and $| \vec{a} | = 8 | \vec{b} |$

$\begin{array}{l} (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \\ a n d | \vec{a} | = 8 | \vec{b} | \\ (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \\ \vec{a} . \vec{a} - \vec{a} . \vec{b} + \vec{b} . \vec{a} - \vec{b} . \vec{b} = 8 \\ {| \vec{a} |}^{2} - {| \vec{b} |}^{2} = 8 \\ {(8 | \vec{b} |)}^{2} - {| \vec{b} |}^{2} = 8 \\ 64 {| \vec{b} |}^{2} - {| \vec{b} |}^{2} = 8 \\ 63 {| \vec{b} |}^{2} = 8 \\ | \vec{b} | =\sqrt8/\sqrt63 (∴ m a g n i t u d e o f a v e c t o r i s n o n - n e g a t i v e) \\ | \vec{b} | = \frac{2√2}{3√7} \\ A n d \\ | \vec{a} | = 8 | \vec{b} | = \frac{2 * 2√2}{3√7} = \frac{1 6√2}{3√7} \end{array}$

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New answer posted

a year ago

Class 12th Vector Algebra

28. Kindly Consider the following

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Vishal Baghel

Contributor-Level 10

Here, each of the given three vector is a unit vector.

$\begin{array}{l} \vec{a} . \vec{b} = \frac{2}{7} * \frac{3}{7} + \frac{3}{7} * (\frac{- 6}{7}) + \frac{6}{7} * \frac{2}{7} \\ = \frac{6}{49} + (\frac{- 18}{49}) + \frac{12}{49} = \frac{6 - 18 + 12}{49} = 0 \\ \vec{b} . \vec{c} = \frac{3}{7} * \frac{6}{7} + (\frac{- 6}{7}) * \frac{2}{7} + \frac{2}{7} * (- \frac{3}{7}) \\ = \frac{18}{49} - \frac{12}{49} + (\frac{- 6}{49}) = \frac{18 - 12 - 6}{49} = 0 \\ \vec{c} . \vec{a} = \frac{6}{7} * \frac{2}{7} + \frac{2}{7} * \frac{3}{7} + (- \frac{3}{7}) * \frac{6}{7} \\ = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = 0 \end{array}$