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…

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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.

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

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