If the foot of the perpendicular from point $(4,3,8)$ on the line $L_1: \frac{x-a}{2}=\frac{y-2}{m}=\frac{z-b}{4}, m \neq 0$ is $(3,5,7)$ then the shortest distance between the line $L_1$ and the line $L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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If the foot of the perpendicular from point

(4,3,8)

on the line

L_1: \frac{x-a}{2}=\frac{y-2}{m}=\frac{z-b}{4}, m \neq 0

(3,5,7)

then the shortest distance between the line

L_1

and the line

L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}

is equal to

View answer and explanation

Correct option: B

\frac{1}{\sqrt{6}}

Option B (

\frac{1}{\sqrt{6}}

) is correct because it represents the shortest distance between the two lines, calculated using the formula involving the vector difference and cross product of direction vectors.

More Questions from SCALAR TRIPLE PRODUCT

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

for some non-zero vector

\vec r,

then the value of

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

is The shortest distance between the lines

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

and

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

is Let

\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}

and

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

Then

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

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

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

equals

a \cdot(a \times b)=

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

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