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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry1 Mark
Question
Assertion (A): The pair of lines given by $\vec{r}=\hat{i}-\hat{j}+\lambda(2 \hat{i}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{k}+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect.
Reason $(R) :$ Two lines intersect each other, if they are not parallel and shortest distance $=0$.
✓

Answer

Here, $\vec{a}_1=\hat{i}-\hat{j}, \vec{b}_1=2 \hat{i}+\hat{k}$
$\vec{a}_2=2 \hat{i}-\hat{k}$ and $\vec{b}_2=\hat{i}+\hat{j}-\hat{k}$
$\because \vec{b}_1 \neq k \vec{b}_2$, for any scalar $k$
$\therefore \quad$ Given lines are not parallel.
Now, $\vec{a}_2-\vec{a}_1=(2 \hat{i}-\hat{k})-(\hat{i}-\hat{j})=\hat{i}+\hat{j}-\hat{k}$
and $\vec{b}_1 \times \vec{b}_2=-\hat{i}+3 \hat{j}+2 \hat{k}$
$\Rightarrow\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{(-1)^2+(3)^2+(2)^2}=\sqrt{1+9+4}=\sqrt{14}$
$\therefore \text { S.D. }=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\left|\frac{(\hat{i}+\hat{j}-\hat{k}) \cdot(-\hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{14}}\right|=0$
Hence, two lines intersect each other.
Two lines intersect each other, if they are not parallel and shortest distance $=0$.

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