Question Number 45809 by rahul 19 last updated on 17/Oct/18 Commented by rahul 19 last updated on 17/Oct/18 $${Calculation}\:{error}…. \\ $$ Commented by MJS last…
Question Number 111347 by Rio Michael last updated on 03/Sep/20 $$\mathrm{convert}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{with}\:\mathrm{cartesian}\:\mathrm{equation}:\:{x}−\mathrm{3}{y}\:+\:\mathrm{2}{z}\:=\:\mathrm{7} \\ $$$$\:\mathrm{into}\:\mathrm{its}\:\mathrm{vector}\:\mathrm{parametric}\:\mathrm{form}. \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 45730 by ajfour last updated on 16/Oct/18 Commented by ajfour last updated on 16/Oct/18 $${If}\:{base}\:{BC}\:{of}\:\bigtriangleup{ABC}\:{has}\:{slipped}\: \\ $$$${to}\:{the}\:{base}\:{edges}\:{of}\:{box}\:{and}\:{its} \\ $$$${sides}\:{AB}\:{and}\:{AC}\:{touch}\:\left({rests}\right. \\ $$$$\left.{against}\right)\:{poimts}\:{P}\:{and}\:{Q} \\ $$$${respectively}\:{of}\:{the}\:{two}\:{upper}…
Question Number 45575 by rahul 19 last updated on 14/Oct/18 Commented by rahul 19 last updated on 14/Oct/18 $${My}\:{doubt}: \\ $$$${If}\:{i}\:{will}\:{take}\:{both}\:{points}\:{on}\:{same}\:{side} \\ $$$${of}\:{plane}\:\left({wrong}\:{method}\right)\:{and}\:{will} \\ $$$${take}\:{points}\:{on}\:{opposite}\:{side}\:{i}\:{will}\:{get}\:…
Question Number 45555 by rahul 19 last updated on 14/Oct/18 $${Prove}\:{that}\:{points}\:\left(\mathrm{4},−\mathrm{1},\mathrm{3}\right)\:\&\:\left(\mathrm{5},−\mathrm{1},\mathrm{4}\right) \\ $$$${lies}\:{on}\:{same}\:{side}\:{of}\:{the}\:{plane}\:{x}+{y}+{z}=\mathrm{7}. \\ $$ Commented by rahul 19 last updated on 14/Oct/18 Answered by…
Question Number 111006 by pete last updated on 01/Sep/20 $$\mathrm{The}\:\mathrm{vectors}\:\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{q}}\:\mathrm{and}\:\boldsymbol{\mathrm{r}}\:\mathrm{are}\:\mathrm{mutially}\:\mathrm{perpendicularwith} \\ $$$$\mid\boldsymbol{\mathrm{q}}\mid=\mathrm{3}\:\mathrm{and}\:\mid\boldsymbol{\mathrm{r}}\mid=\sqrt{\mathrm{5}.\mathrm{4}\:}\:.\mathrm{If}\:\mathrm{X}=\:\mathrm{7}\boldsymbol{\mathrm{p}}+\mathrm{5}\boldsymbol{\mathrm{q}}+\mathrm{7}\boldsymbol{\mathrm{r}}\:\mathrm{and} \\ $$$$\mathrm{Y}=\mathrm{2}\boldsymbol{\mathrm{p}}+\mathrm{3}\boldsymbol{\mathrm{q}}−\mathrm{5}\boldsymbol{\mathrm{r}}\:\mathrm{are}\:\mathrm{perpendicular},\:\mathrm{find}\mid\boldsymbol{\mathrm{p}}\mid. \\ $$ Commented by kaivan.ahmadi last updated on 01/Sep/20 $${p}.{q}={p}.{r}={q}.{r}=\mathrm{0} \\…
Question Number 45417 by rahul 19 last updated on 12/Oct/18 $${Find}\:{image}\:{of}\:{plane}\:{x}−{y}+{z}−\mathrm{3}=\mathrm{0}\:{in}\: \\ $$$${plane}\:\mathrm{2}{x}+{y}−{z}+\mathrm{4}=\mathrm{0}\:? \\ $$ Answered by MrW3 last updated on 13/Oct/18 $${since}\:{both}\:{planes}\:{are}\:{perpendicular} \\ $$$${to}\:{each}\:{other},\:{the}\:{image}\:{is}\:{itself},\:{i}.{e}.…
Question Number 45259 by rahul 19 last updated on 11/Oct/18 $${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 by rahul 19 last updated on 11/Oct/18…
Question Number 44848 by pieroo last updated on 05/Oct/18 $$\left(\mathrm{1}\right)\:\mathrm{If}\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}=\mathrm{a}_{\mathrm{1i}} +\mathrm{a}_{\mathrm{2j}} +\mathrm{a}_{\mathrm{3k}} \:\mathrm{and}\:\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\mathrm{b}_{\mathrm{1i}} +\mathrm{b}_{\mathrm{2j}} +\mathrm{b}_{\mathrm{3k}} \:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{\mathrm{a}_{\mathrm{1}} }&{\mathrm{a}_{\mathrm{2}} }&{\mathrm{a}_{\mathrm{3}} }\\{\mathrm{b}_{\mathrm{1}}…
Question Number 44035 by rahul 19 last updated on 20/Sep/18 Commented by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18 $$\left(\overset{\rightarrow} {{x}}.\overset{\rightarrow} {{z}}\right)\overset{\rightarrow} {{y}}−\left(\overset{\rightarrow} {{x}}.\overset{\rightarrow} {{y}}\right)\overset{\rightarrow} {{z}}=\overset{\rightarrow} {{a}}…