Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

Statement -1: If a is an integer, then the straight lines $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + λ(a \hat{i} + 2\hat{j}+3 \hat{k})$

and $\vec{r}= 2\hat{i} + 3\hat{j} + \hat{k} + \mu (3\hat{i} + a \hat{j} + 2\hat{k})$ intersect at a point for a = -5.

Statement-2 : Two straight lines intersect if the shortest distance  between them is zero.

Options:

Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1.

Statement 1 is True, Statement 2 is False.

Statement 1 is False, Statement 2 is True.

Correct Answer:

Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Explanation:

Clearly, statement-2 is true.

Given lines pass through $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{a}_2 = 2\hat{i} + 3\hat{j} + 3\hat{k}$ respectively and are parallel to vectors $\vec{b}_1 = a\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_2= 3 \hat{i} + a\hat{j} + 2\hat{k}$ respectively.

If given lines intersect, then, S.D. = 0.

$⇒ (\vec{b}_1 × \vec{b}_2). (\vec{a}_1 - \vec{a} - \vec{a}_2)= 0.$

$⇒ \begin{vmatrix}a & 2 & 3\\3 & a & 2\\1 & 1 & -2\end{vmatrix}= 0 $

$⇒ a(-2a-2) - 2(-6-2) + 3(3-a)= 0 $

$⇒ 2a^2 + 5a - 25 = 0 ⇒(a + 5) (2a - 5) = 0 ⇒ a = -5, 5/2$