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Find the vector equation of the line passing through the point (1,2,-4) and perpendicular to the two lines: \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
CBSE | Class 12 | Maths | Miscellaneous Exercise | NCERT | Three Dimensional Geometry
Find the vector equation of the line passing through the point (1,2,-4) and perpendicular to the two lines: \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}

Solution:

The vector eq. of a line passing through a point with position vector \vec{a} and parallel to a vector \overrightarrow{\mathrm{b}} is \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}
Given: the line passes through (1,2,-4)

Therefore,
\overrightarrow{\mathrm{a}}=1 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}
Also given that, line is parallel to both planes.
Therefore it can be said that the line is perpendicular to normal of both planes.
That is \vec{b} is perpendicular to normal of both planes.

It is known that
\vec{a} \times \vec{b} is perpendicular to both \vec{a} \& \vec{b}
Therefore, \overrightarrow{\mathrm{b}} is cross product of normal of planes
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}

\begin{array}{l} \text { Required Normal }=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \\ =\hat{i}[(-16)(-5)-8(7)]-\hat{j}[3(-5)-3(7)]+\hat{k}[3(8)-3(-16)] \\ =\hat{i}[80-56]-\hat{j}[-15-21]+\hat{k}[24+48] \\ =24 \hat{i}+36 \hat{j}+72 \hat{k} \end{array}

So,
\overrightarrow{\mathrm{b}}=24 \hat{\mathrm{i}}+36 \hat{\mathrm{j}}+72 \hat{\mathrm{k}}
Substituting the value of \vec{a} \& \vec{b} in the formula, we obtain

\begin{aligned} \vec{r} =\vec{a}+\lambda \vec{b} \\ =(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(24 \hat{i}+36 \hat{j}+72 \hat{k}) \\ =(1 \hat{i}+2 \hat{j}-4 \hat{k})+12 \lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ =(1 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \end{aligned}

As a result, the equation of the line is
\vec{r}=(1 \hat{\boldsymbol{i}}+2 \hat{\boldsymbol{j}}-4 \hat{\boldsymbol{k}})+\lambda(2 \hat{\boldsymbol{i}}+3 \hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}})

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