# Vector Algebra - Class 12 - Mathematics

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## Extra Questions - Vector Algebra | NCERT | Mathematics | Class 12

The position vector of the midpoint of joining the points $(2,-1,3)$ and $(4,3,-5)$ is:

(A) $3\hat{i} + \hat{j} - \hat{k}$

(B) $\hat{i} + \hat{j} + \hat{k}$

(C) $-3\hat{i} - \hat{j} - \hat{k}$

(D) $3\hat{i} + 2\hat{j} - 7\hat{k}$

The correct answer is **Option (A)**:

$$ 3\hat{i} + \hat{j} - \hat{k} $$

To solve this, let's define the position vectors:

For the point $(2,-1,3)$, the position vector is $\overrightarrow{OA} = 2\hat{i} - \hat{j} + 3\hat{k}$.

For the point $(4,3,-5)$, the position vector is $\overrightarrow{OB} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.

We can find the position vector of the midpoint $\overrightarrow{OM}$ by averaging these two vectors:

$$ \overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{(2\hat{i} - \hat{j} + 3\hat{k}) + (4\hat{i} + 3\hat{j} - 5\hat{k})}{2} $$

Breaking down the summation, we have:

$$ \overrightarrow{OM} = \frac{2\hat{i} + 4\hat{i} - \hat{j} + 3\hat{j} + 3\hat{k} - 5\hat{k}}{2} = \frac{6\hat{i} + 2\hat{j} - 2\hat{k}}{2} = 3\hat{i} + \hat{j} - \hat{k} $$

This simplifies to the correct answer, which is:

$$ \overrightarrow{OM} = 3\hat{i} + \hat{j} - \hat{k} $$

Hence, **Option (A)** is indeed the right choice.

Let $\vec{a}, \vec{b}$, and $\vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} \times \vec{c})$. If $\vec{b}$ is not parallel to $\vec{c}$, then the angle between $\vec{a}$ and $\vec{b}$ is:

(A) $\frac{\pi}{2}$

B) $\frac{2 \pi}{3}$ (C) $\frac{5 \pi}{6}$

D) $\frac{3 \pi}{4}$

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Convert the vector rvector is equal to 3i cap + 2j cap into the unit vector.

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Find the equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i}-3 \hat{j}+4 \hat{k}$.

A) $2x + 3y + 4z = 6$

B) $3x- 4y + 2z = 6$

C) $2x - 3y + 4z = 6$

D) $2x - 3y - 4z = 6$