Question Number 45259 by rahul 19 last updated on 11/Oct/18 | ||
$${Find}\:\:{distance}\:{of}\:{point}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{from} \\ $$ $${the}\:{line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:\& \\ $$ $$\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right)\:? \\ $$ | ||
Commented byrahul 19 last updated on 11/Oct/18 | ||
$${Another}\:{method}: \\ $$ $${Let}\:{P}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:,\:{Q}\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:,\:{R}\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$ $${and}\:{M}\:{be}\:{foot}\:{of}\:{perpendicular}... \\ $$ $${Now}\:{in}\:\Delta{PQR}, \\ $$ $${area}\:{of}\:\Delta=\:\frac{\mathrm{1}}{\mathrm{2}}\mid{PQ}×{QR}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{PM}×{QR}\mid \\ $$ $$\Rightarrow\:\mid\left({i}+\mathrm{2}{j}+\mathrm{3}{k}\right)×\left(\mathrm{3}{i}+{j}+{k}\right)\mid={PM}×\sqrt{\mathrm{11}} \\ $$ $$\Rightarrow\:{PM}=\:\sqrt{\frac{\mathrm{90}}{\mathrm{11}}} \\ $$ | ||
Answered by ajfour last updated on 11/Oct/18 | ||