Ncert Solutions Maths class 12th NCERT Vector Algebra

What is an orthogonal triad in Vector Algebra and is it covered in Shiksha's NCERT Solutions?

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5 months ago

As per the NCERT Textbooks, An orthogonal triad in vector algebra is a 3D cordinate point system where three vectors that are mutually perpendicular to each other, meaning the angle between each pair of vectors is 90°. These vectors are typically non-zero, linearly independent, and often form the basis of 3D space like the i? , j? , k? unit vectors in the Cartesian coordinate system.

Yes, Shiksha's NCERT Solutions for Class 12 Vector Algebra do cover the concept of orthogonal triads, We have used this concept specially while explaining dot product, cross product, and geometric interpretation of vectors.

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If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and $\vec{b}$ and $\vec{c}$ are non-collinear. $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.

alok kumar singh

$\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$

⇒ $\vec{a} + 5 \vec{b}$ = $\vec{c}$ …(1)

$\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$

⇒ $\vec{b} + 6 \vec{c} = μ \vec{a}$ …(2)

From (1) and (2)

$\vec{b} + 6 \vec{c} = μ (λ \vec{c} - 5 \vec{b})$

-> $(1 + 5 μ) \vec{b} + (6 - λ μ) \vec{c} = 0$

$? \vec{b}$ and $\vec{c}$ are non-collinear

-> 1 + 5m = 0 $μ = \frac{- 1}{5}$ and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\vec{b} = 6 \vec{c} = \frac{- 1}{5} \vec{a}$

$5 \vec{b} + 30 \vec{c} = - \vec{a}$

$\begin{matrix} \vec{a} + 5 \vec{b} + 30 \vec{c} = 0 & \vec{a} + α \vec{b} + β \vec{c} = 0 \end{matrix}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35

Let $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} a n d \vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} = \vec{c} \times \vec{a} a n d \vec{r} . \vec{b} = 0,$ then $\vec{r} . \vec{a}$ is equal to………………

Vishal Baghel

$\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j}, \vec{c} = \hat{i} - \hat{j} - \hat{k}$

$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$

$\vec{r} = \vec{c} + λ \vec{a}$

Now, $0 = \vec{b} . \vec{c} + λ \vec{a} . \vec{b} a s \vec{r} . \vec{b} = 0$

$λ = \frac{- \vec{b} . \vec{c}}{\vec{a} . \vec{b}} = 2$

$∴ \vec{r} . \vec{a} = \vec{a} . \vec{c} + 2 a^{2} = 12$

If vectors ${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} a n d {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is:

alok kumar singh

${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} & {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$

given ${\vec{a}}_{1} & {\vec{a}}_{2}$ are collinear then ${\vec{a}}_{1} = λ {\vec{a}}_{2}$

$\Rightarrow (x \hat{i} - \hat{j} + \hat{k}) = λ (\hat{i} + y \hat{j} + z \hat{k})$

Since $\hat{i}, \hat{j} & \hat{k}$ are not collinear so

$S o x \hat{i} + y \hat{j} + z \hat{k} = λ \hat{i} - \frac{1}{λ} \hat{j} + \frac{1}{λ} \hat{k}$

Hence possible unit vector parallel to it be $\frac{1}{\sqrt[]{3}} (\hat{i} - \hat{j} + \hat{k})$ for $λ$ =

Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k} a n d \vec{c}$ be a vector such that $\vec{a} + (\vec{b} \times \vec{c}) = \vec{0} a n d \vec{b} . \vec{c} = 5 .$ Then, the value of $3 (\vec{c} . \vec{a})$ is equal to……….

Vishal Baghel

Data contradiction.

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

Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If a is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is:

Vishal Baghel

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$ $\frac{}{}$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

= 2

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