RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS3 Marks
Question
The cartesian equation of a line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$ Write its vector form.
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Answer
Given: The Cartesian equation of the line is
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}=\lambda\ (\text{say})$
$\Rightarrow\ \ \text{x}-5=3\lambda,\ \text{y}+4=7\lambda,\ \text{z}-6=2\lambda$
$\Rightarrow\ \ \text{x}=5+3\lambda,\ \text{y}=-4+7\lambda,\ \text{z}=6+2\lambda$
General equation for the required line is
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Putting the values of x, y, z in this equation,
$\vec{\text{r}}=(5+3\lambda)\hat{\text{i}}+(-4+7\lambda)\hat{\text{j}}+(6+2\lambda)\hat{\text{k}}$
$=5\hat{\text{i}}+3\lambda\hat{\text{i}}-4\hat{\text{j}}+7\lambda\hat{\text{j}}+6\hat{\text{k}}+2\lambda\hat{\text{k}}$
$\Rightarrow\ \ \ \ \vec{\text{r}}=\Big(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}\Big)+\lambda\Big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\Big)\ \ \ \ \Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
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