Tags Class 12th

Class 12th

Get insights from 11.8k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

11.8k

Questions

Discussions

Active Users

Followers

New answer posted

8 months ago

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 $\vec{b}$

0 Follower 9 Views

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.

0 0

New answer posted

8 months 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

0 Follower 43 Views

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.

0 0

New answer posted

8 months ago

Class 12th Vector Algebra

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

8 months 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.

0 Follower 6 Views

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

0 0

New answer posted

8 months ago

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.

0 Follower 5 Views

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.

0 0

New answer posted

8 months ago

Class 12th Vector Algebra

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

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

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

0 0

New answer posted

8 months ago

Class 12th Vector Algebra

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

0 0

New answer posted

8 months ago

Class 12th Vector Algebra

28. Kindly Consider the following

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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