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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Show that the lines
$\vec{\text{r}}=3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ are intersecting. Hence, find their point of intersection.
✓

Answer

The position vectors of two arbitrary points on the given lines are
$3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(3+\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(2\lambda-4)\hat{\text{k}}$
$5\hat{\text{i}}-2\hat{\text{j}}+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
$=(5+3\mu)\hat{\text{i}}+(-2+2\mu)\hat{\text{j}}+6\mu\hat{\text{k}}$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$(3+\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(2\lambda-4)\hat{\text{k}}$
$=(5+3\mu)\hat{\text{i}}+(-2+2\mu)\hat{\text{j}}6\mu\hat{\text{k}}$
Equation the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$3+\lambda=5+3\mu\dots(1)$
$2+2\lambda=-2+2\mu\dots(2)$
$2\lambda-4=6\mu\dots(3)$
Solving (1) and (2), we get
$\lambda=-4,\mu=-2.$
Substituting the values $\lambda=-4$ and $\mu=-2$ in (3), we get
$\text{LHS}=2\lambda-4$
$=2(-4)-4$
$=-12\text{ RHS}=6\mu$
$=6(-2)$
$=-12$
$\Rightarrow\text{LHS}=\text{RHS}$
Since $\lambda=-4$ and $\mu=-2$ satisfy (3), the lines intersect.
Substituting $\mu=-2$ in the second line, we get $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{i}}-4\hat{\text{j}}-12\hat{\text{k}}=-\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$ the position vector of the point of intersection.
Thus, the coordinates of the points of intersection are (-1, -6, -12).

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