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Question

Question: The vector equation of the line joining the points \(\mathbf { i } - 2 \mathbf { j } + \mathbf { k ...

The vector equation of the line joining the points i2j+k\mathbf { i } - 2 \mathbf { j } + \mathbf { k } and 2j+3k- 2 \mathbf { j } + 3 \mathbf { k } is

A

r=t(i+j+k)\mathbf { r } = t ( \mathbf { i } + \mathbf { j } + \mathbf { k } )

B

r=t1(i2j+k)+t2(3k2j)\mathbf { r } = t _ { 1 } ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) + t _ { 2 } ( 3 \mathbf { k } - 2 \mathbf { j } )

C

r=(i2j+k)+t(2ki)\mathbf { r } = ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) + t ( 2 \mathbf { k } - \mathbf { i } )

D

r=t(2ki)\mathbf { r } = t ( 2 \mathbf { k } - \mathbf { i } )

Answer

r=(i2j+k)+t(2ki)\mathbf { r } = ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) + t ( 2 \mathbf { k } - \mathbf { i } )

Explanation

Solution

Vector equation of a straight line passing through two points a and b is, r=a+t(ba)\mathbf { r } = \mathbf { a } + t ( \mathbf { b } - \mathbf { a } )

r=(i2j+k)+t(2ki)\mathbf { r } = ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) + t ( 2 \mathbf { k } - \mathbf { i } )