Class CBSE Class 12 Mathematics Vector Algebra Q #1479
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
The scalar product of the vector $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$ with a unit vector along sum of vectors $\vec{b}=2\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{c}=\lambda\hat{i}-2\hat{j}-3\hat{k}$ is equal to 1. Find the value of $\lambda$.

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Find the sum of vectors $\vec{b}$ and $\vec{c}$

We are given $\vec{b}=2\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{c}=\lambda\hat{i}-2\hat{j}-3\hat{k}$. So, $\vec{b}+\vec{c} = (2+\lambda)\hat{i} + (-4-2)\hat{j} + (5-3)\hat{k} = (2+\lambda)\hat{i} -6\hat{j} + 2\hat{k}$.

Step 2: Find the unit vector along $\vec{b}+\vec{c}$

Let $\hat{u}$ be the unit vector along $\vec{b}+\vec{c}$. Then, $\hat{u} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}$. First, we find the magnitude of $\vec{b}+\vec{c}$: $|\vec{b}+\vec{c}| = \sqrt{(2+\lambda)^2 + (-6)^2 + (2)^2} = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40}$. So, $\hat{u} = \frac{(2+\lambda)\hat{i} -6\hat{j} + 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}}$.

Step 3: Find the scalar product of $\vec{a}$ and $\hat{u}$

We are given $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$. The scalar product of $\vec{a}$ and $\hat{u}$ is given by $\vec{a} \cdot \hat{u} = 1$. So, $(\hat{i}-\hat{j}+2\hat{k}) \cdot \frac{(2+\lambda)\hat{i} -6\hat{j} + 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}} = 1$. $\frac{(1)(2+\lambda) + (-1)(-6) + (2)(2)}{\sqrt{(2+\lambda)^2 + 40}} = 1$. $\frac{2+\lambda + 6 + 4}{\sqrt{(2+\lambda)^2 + 40}} = 1$. $\frac{\lambda + 12}{\sqrt{(2+\lambda)^2 + 40}} = 1$.

Step 4: Solve for $\lambda$

Squaring both sides, we get: $(\lambda + 12)^2 = (2+\lambda)^2 + 40$. $\lambda^2 + 24\lambda + 144 = \lambda^2 + 4\lambda + 4 + 40$. $24\lambda + 144 = 4\lambda + 44$. $20\lambda = -100$. $\lambda = -5$.

Final Answer: -5

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because the student needs to recall the formula for the scalar product of two vectors and the definition of a unit vector to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a series of steps, including finding the sum of vectors, calculating the unit vector, and using the scalar product formula.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of vector algebra concepts as covered in the textbook.

More from this Chapter

SUBJECTIVE
(i) Complete the given figure to explain their entire movement plan along the respective vectors. (ii) Find vectors $\vec{AC}$ and $\vec{BC}$. (iii) (a) If $\vec{a} \cdot \vec{b} = 1$, distance of O to A is 1 km and that from O to B is 2 km, then find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$. Also, find $| \vec{a} \times \vec{b}|$. OR (iii) (b) If $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$, then find a unit vector perpendicular to $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$.
MCQ_SINGLE
If vector \(\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}\) and vector \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), then which of the following is correct ?
MCQ_SINGLE
A student tries to tie ropes, parallel to each other from one end of the wall to the other. If one rope is along the vector \(3\hat{i}+15\hat{j}+6\hat{k}\) and the other is along the vector \(2\hat{i}+10\hat{j}+\lambda\hat{k}\), then the value of \(\lambda\) is :
VSA
If $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}$, find the value of $(\vec{r}\times\hat{j})\cdot(\vec{r}\times\hat{k})-12$
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 $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
View All Questions