Class CBSE Class 12 Mathematics Vector Algebra Q #1361
KNOWLEDGE BASED
REMEMBER
2 Marks 2025 AISSCE(Board Exam) VSA
Find a vector of magnitude 21 units in the direction opposite to that of $\vec{AB}$ where A and B are the points $A(2,1,3)$ and $B(8,-1,0)$ respectively.
Prev Next

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Find the vector $\vec{AB}$

The vector $\vec{AB}$ is given by the difference of the position vectors of points B and A.

$\vec{AB} = \vec{OB} - \vec{OA} = (8\hat{i} - \hat{j} + 0\hat{k}) - (2\hat{i} + \hat{j} + 3\hat{k}) = (8-2)\hat{i} + (-1-1)\hat{j} + (0-3)\hat{k} = 6\hat{i} - 2\hat{j} - 3\hat{k}$

Step 2: Find the vector in the opposite direction of $\vec{AB}$

The vector in the opposite direction of $\vec{AB}$ is $-\vec{AB}$.

$-\vec{AB} = - (6\hat{i} - 2\hat{j} - 3\hat{k}) = -6\hat{i} + 2\hat{j} + 3\hat{k}$

Step 3: Find the magnitude of $-\vec{AB}$

The magnitude of $-\vec{AB}$ is given by:

$|-\vec{AB}| = \sqrt{(-6)^2 + (2)^2 + (3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$

Step 4: Find the unit vector in the direction of $-\vec{AB}$

The unit vector in the direction of $-\vec{AB}$ is given by:

$\hat{u} = \frac{-\vec{AB}}{|-\vec{AB}|} = \frac{-6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = -\frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k}$

Step 5: Find the vector of magnitude 21 in the direction of $-\vec{AB}$

The vector of magnitude 21 in the direction of $-\vec{AB}$ is given by:

$\vec{v} = 21 \hat{u} = 21 \left( -\frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k} \right) = -18\hat{i} + 6\hat{j} + 9\hat{k}$

Final Answer: $-18\hat{i} + 6\hat{j} + 9\hat{k}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the formula for finding a vector between two points and the concept of unit vectors and scalar multiplication.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a series of steps to arrive at the solution, such as finding the vector between two points, finding the unit vector, and then scaling it to the desired magnitude.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests a standard procedure covered in the textbook. The question assesses the student's ability to apply vector algebra concepts.

More from this Chapter

MCQ_SINGLE
The respective values of \(|\vec{a}|\) and \(|\vec{b}|\), if given \((\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})=512\) and \(|\vec{a}|=3|\vec{b}|\), are:
MCQ_SINGLE
Let \(|\vec{a}|=5\) and \(-2\le\lambda\le1\). Then, the range of \(|\lambda\vec{a}|\) is:
VSA
A vector $\vec{a}$ makes equal angles with all the three axes. If the magnitude of the vector is $5\sqrt{3}$ units, then find $\vec{a}$.
VSA
(a) If the vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 3$, $|\vec{b}| = \frac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector, then find the angle between $\vec{a}$ and $\vec{b}$. OR(b) Find the area of a parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
SA
The position vectors of vertices of $\Delta$ ABC are $A(2\hat{i}-\hat{j}+\hat{k}),$ $B(\hat{i}-3\hat{j}-5\hat{k})$ and $C(3\hat{i}-4\hat{j}-4\hat{k})$ Find all the angles of $\Delta$ AВС.
View All Questions