What are the important topics in Class 12 Vector Algebra?

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5 months ago

Class 12 Vector Algebra consists of several important topics for not only class 12 board exams but also varous competitive exams. Check the several key topics of the chapter below;

Types of Vectors: Unit, Position, Co-initial, collinear, and coplanar vectors
Operations on Vectors: Addition, subtraction, and Dot & Cross multiplication.
Properties of cross-product: Geometric interpretation, Area of a parallelogram, Triangle

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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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