Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #879
KNOWLEDGE BASED
APPLY
3 Marks 2023 SA
Find the distance between the lines: →r = (î + 2î - 4ê) + λ(2î + 3î + 6ê); →r = (3î + 3î - 5ê) + μ(4î + 6î + 12ê)
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

Step 1: Check if the lines are parallel.

The direction vectors of the lines are b1 = 2î + 3ĵ + 6k and b2 = 4î + 6ĵ + 12k.

Notice that b2 = 2 * b1, which means the lines are parallel.

Step 2: Find a point on each line.

Point A on the first line: (1, 2, -4)

Point B on the second line: (3, 3, -5)

Step 3: Find the vector connecting the two points.

a = AB = (3-1)î + (3-2)ĵ + (-5-(-4))k = 2î + ĵ - k

Step 4: Calculate the cross product of the direction vector and the vector connecting the points.

b1 x a = (2î + 3ĵ + 6k) x (2î + ĵ - k) =
î(3*(-1) - 6*1) - ĵ(2*(-1) - 6*2) + k(2*1 - 3*2) = -9î + 14ĵ - 4k

Step 5: Find the magnitude of the cross product.

|b1 x a| = √((-9)^2 + (14)^2 + (-4)^2) = √(81 + 196 + 16) = √293

Step 6: Find the magnitude of the direction vector.

|b1| = √(2^2 + 3^2 + 6^2) = √(4 + 9 + 36) = √49 = 7

Step 7: Calculate the distance between the parallel lines.

Distance = |b1 x a| / |b1| = √293 / 7

Correct Answer: √293 / 7

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 finding the distance between two skew lines or parallel lines after verifying their relationship.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure or algorithm to calculate the distance between the lines. This involves identifying the direction vectors, checking for parallelism, and then applying the appropriate formula.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of formulas and procedures related to 3D geometry as covered in the textbook.

More from this Chapter

MCQ_SINGLE
The equation of a line parallel to the vector \(3\hat{i}+\hat{j}+2\hat{k}\) and passing through the point \((4, -3, 7)\) is:
MCQ_SINGLE
The value of $\lambda$ for which the angle between the lines $\vec{r}=\hat{i}+\hat{j}+\hat{k}+p(2\hat{i}+\hat{j}+2\hat{k})$ and $\vec{r}=(1+q)\hat{i}+(1+q\lambda)\hat{j}+(1+q)\hat{k}$ is $\frac{\pi}{2}$ :
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 $\beta$.
MCQ_SINGLE
The line \(x=1+5\mu\), \(y=-5+\mu\), \(z=-6-3\mu\) passes through which of the following point ?
LA
Find the value of $b$ so that the lines $\frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ are intersecting lines. Also, find the point of intersection of these given lines.
View All Questions