Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector i -2 j +3 k . Reduce the corresponding equation in cartesian form.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + b . Here, a = i +2 j +3 k b = i -2 j +3 k vector equation of the required line is r = ( i +2 j +3 k )+ ( i +2 j +3 k ) Here, is a parameter. Reducing (1) to cartesian form, we get x i +y i +z k = ( i +2 j +3 k )+ ( i -2 j +3 k ) [putting r =x i +y j +z k in (1) ] i +y j +z k =(1+ ) i +(2-2 ) j +(3+3 ) k Comparing the cofficients of i , j and k , we get x=1+ ,y=2-2 ,z=3+3 -1= , y-2 -2 = , z-3 3 = x-1 1 = y-2 -2 = z-3 3 = Hence, the cartesian form of (1) is x-1 1 = y-2 -2 = z-3 3

Find the vector equation for the line which passes through the po…

Download App

Question

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}.$ Reduce the corresponding equation in cartesian form.

✓

Answer

We know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
vector equation of the required line is
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
Reducing (1) to cartesian form, we get
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{i}}+\text{z}\hat{\text{k}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)$ $\big[\text{putting }\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}} \text{ in }(1)\big]$
$\Rightarrow\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(1+\lambda)\hat{\text{i}}+(2-2\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}$
Comparing the cofficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$\text{x}=1+\lambda,\text{y}=2-2\lambda,\text{z}=3+3\lambda$
$\Rightarrow\text{x}-1=\lambda,\frac{\text{y}-2}{-2}=\lambda,\frac{\text{z}-3}{3}=\lambda$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-3}{3}=\lambda$
Hence, the cartesian form of (1) is
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-3}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(5 Marks)" from Straight line in space→Straight line in space — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Straight line in space — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions