Question bank/JEE Main/Mathematics

The expression in the vector form for the point
[IMAGE] of intersection of the plane
$\overrightarrow{r}.\overrightarrow{n} = d$ and the perpendicular line
$\overrightarrow{r} = \overrightarrow{r_{o}} + t\overrightarrow{n}$
where t is a parameter given by.
$\overrightarrow{r} = \overrightarrow{r_{o}} + t\overrightarrow{n}$ tgk¡
t izkpy izfrPNsnu

A${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$B${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} - \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$C${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n} - d}{|\overset{\rightarrow}{n}|} \right)\overset{\rightarrow}{n}$D${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{|\overset{\rightarrow}{n}|} \right)\overset{\rightarrow}{n}$

Show answer and solution▾

$\Rightarrow {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n} + t(\overset{\rightarrow}{n})^{2} = d$
$t = \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{(\overset{\rightarrow}{n})^{2}}$
$\therefore{\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx