Question

Vector equation of the line 6x – 2 = 3y + 1 = 2z – 2 is :

• A. $\overrightarrow { \mathbf { r } }$ = $\hat { \mathbf { i } }$ – + $\hat { \mathbf { k } }$ + l ( $\hat { \mathbf { i } }$ – $\hat { \mathrm { j } }$ + $\hat { \mathbf { k } }$ )
• B. $\overrightarrow { \mathbf { r } }$ = $\hat { \mathbf { i } }$ + 2 + $\hat { \mathbf { k } }$ + l
• C. = – + $\hat { \mathbf { k } }$ + l( – + )
• D. = $\frac { 1 } { 3 }$( – + 3) + l( + 2 + 3)

Answer 1

Answer: (D)

= $\frac { 1 } { 3 }$( – + 3) + l( + 2 + 3)

Answer 2

Here,

$\frac { x - 1 / 3 } { 1 / 6 }$ = $\frac { y + 1 / 3 } { 1 / 3 }$ = $\frac { z - 1 } { 1 / 2 }$ …(i)

Line $\overrightarrow { \mathbf { r } }$ = + l can be written as

$\frac { x - a _ { 1 } } { b _ { 1 } }$ = $\frac { y - a _ { 2 } } { b _ { 2 } }$ = $\frac { z - a _ { 3 } } { b _ { 3 } }$

Where = a1­ $\hat { \mathbf { i } }$ + a2 + a3 $\hat { \mathbf { k } }$

and $\overrightarrow { \mathrm { b } }$ = b1 + b2 + a3 $\hat { \mathbf { k } }$

so the equation (i) can be written as :

+ l( + 2 + 3)