CBSE Class 12 Mathematics Three Dimensional Geometry Q #885
KNOWLEDGE BASED
APPLY
2 Marks 2023 VSA
If the angle between the lines $\frac{x-5}{\alpha}=\frac{y+2}{-5}=\frac{z+\frac{24}{5}}{\beta}$ and $\frac{x}{1}=\frac{y}{0}=\frac{z}{1}$ is $\frac{\pi}{4}$, find the relation between $\alpha$ and $\beta$.

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

  1. The direction ratios of the first line are $\alpha, -5, \beta$ and the direction ratios of the second line are $1, 0, 1$.

  2. The angle between the lines is given by $\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$.

  3. Here, $\theta = \frac{\pi}{4}$, so $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

  4. Substituting the direction ratios, we get $\frac{1}{\sqrt{2}} = \frac{\alpha(1) + (-5)(0) + \beta(1)}{\sqrt{\alpha^2 + (-5)^2 + \beta^2}\sqrt{1^2 + 0^2 + 1^2}}$.

  5. Simplifying, $\frac{1}{\sqrt{2}} = \frac{\alpha + \beta}{\sqrt{\alpha^2 + 25 + \beta^2}\sqrt{2}}$.

  6. Further simplification gives $1 = \frac{\alpha + \beta}{\sqrt{\alpha^2 + 25 + \beta^2}}$.

  7. Squaring both sides, we get $1 = \frac{(\alpha + \beta)^2}{\alpha^2 + 25 + \beta^2}$.

  8. Therefore, $\alpha^2 + 25 + \beta^2 = (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$.

  9. This simplifies to $25 = 2\alpha\beta$.

  10. Hence, the relation between $\alpha$ and $\beta$ is $\alpha\beta = \frac{25}{2}$.

Correct Answer: $\alpha\beta = \frac{25}{2}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the formula for the angle between two lines in 3D space, given their direction ratios, to find the relation between the unknowns α and β.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concept of direction ratios and the formula for the angle between two lines in 3D space.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of the formula for the angle between two lines in 3D geometry, a standard topic covered in the textbook.

More from this Chapter

SA
Find the coordinates of the foot of the perpendicular drawn from the point $P(0, 2, 3)$ to the line:$$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$ OR (b) Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ satisfy the condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Evaluate the quantity $\mu = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$, if $|\vec{a}| = 3$, $|\vec{b}| = 4$, and $|\vec{c}| = 2$.
LA
Two vertices of the parallelogram ABCD are given as $A(-1,2,1)$ and $B(1,-2,5)$. If the equation of the line passing through C and D is $\frac{x-4}{1}=\frac{y+7}{-2}=\frac{z-8}{2}$ then find the distance between sides AB and CD. Hence, find the area of parallelogram ABCD.
SA
Find the distance between the lines:$$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$$$\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$$
LA
Find the co-ordinates of the foot of the perpendicular drawn from the point (2, 3, -8) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ Also, find the perpendicular distance of the given point from the line.
LA
Find the vector and the Cartesian equations of a line passing through the point (1,2,-4) and parallel to the line joining the points A(3,3,-5) and B(1,0,-11). Hence, find the distance between the two lines. OR Find the equations of the line passing through the points A(1,2,3) and B(3,5,9). Hence, find the coordinates of the points on this line which are at a distance of 14 units from point B.
View All Questions