The shortest distance between the straight lines through the points ${A_1} = \left( {6,2,2} \right){\rm{\;}}$ and ${\rm{\;}}{A_2} = \left( { - 4,0, - 1} \right)$ in the directions of $\left( {1, - 2,2} \right)$ and $\left( {3, - 2, - 2} \right)$ is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Application +4 | -1

The shortest distance between the straight lines through the points

{A_1} = \left( {6,2,2} \right)\mathrm{\;}

and

\mathrm{\;}{A_2} = \left( { - 4,0, - 1} \right)

in the directions of

\left( {1, - 2,2} \right)

and

\left( {3, - 2, - 2} \right)

View answer and explanation

Correct option: D

The correct answer is D (9) because it is calculated using the shortest distance formula between two skew lines.

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}