If the shortest distance between the straight lines $3(x - 1) = 6(y - 2) = 2(z - 1)$ and $4(x - 2) = 2(y - \lambda ) = (z - 3),\,\,\lambda \in R$ is $\frac{1}{{\sqrt {38} }}$, then the integral value of $\lambda $ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

If the shortest distance between the straight lines

3(x - 1) = 6(y - 2) = 2(z - 1)

and

4(x - 2) = 2(y - \lambda ) = (z - 3),\,\,\lambda \in R

\frac{1}{{\sqrt {38} }}

, then the integral value of

\lambda

is equal to

View answer and explanation

Correct option: A

Option A (3) is correct because the shortest distance between the lines is given as

\frac{1}{\sqrt{38}}

, and solving for

\lambda

using the formula for shortest distance yields

\lambda = 3

Practice All Chapter MCQs →

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

\vec r,

\left[ {\vec a\vec b\vec c} \right],

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

\vec{a}, \vec{b}

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

a \cdot(a \times b)=

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}