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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Prove that the line $\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)$ and $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$ intersect and find their point of intersection.
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Answer

The position vectors of two arbitrary points on the given lines are
$\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)=(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}$
$\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
Equating the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$1+3\lambda=4+2\mu\dots(1)$
$1-\lambda=0\dots(2)$
$3\mu-1=-1\dots(3)$
Solving (2) and (3), we get
$\lambda=1$
$\mu=0$
Substituting the valuse $\lambda=1$ and $\mu=0$ in (1), we get
$\text{LHS}=1+3\lambda$
$=1+3(1)$
$=4$
$\text{RHS}=4+2\mu$
$=4+2(0)$
$=4$
$\Rightarrow\text{LHS}=\text{RHS}$
Since $\lambda=1$ and $\mu=0$ satisfy (3), the given lines intersect.
Substituting $\mu=0$ in the second line, we get $\vec{\text{r}}=4\hat{\text{i}}+0\hat{\text{j}}-\hat{\text{k}}$ as the position vector of the point of intersection.
Thus, the coordinates of the point of intersection are (4, 0, -1).

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