Question Number 45898 by rahul 19 last updated on 18/Oct/18 Commented by rahul 19 last updated on 18/Oct/18 $${Let}\:{v}\:=\:{xi}+{yj}+{zk}\:.{Since}\:{a}\:\&{b}\:{are}\:{not}\:{collinear} \\ $$$$\Rightarrow\:{v}\:=\:\lambda{a}+\mu{b}\:\Rightarrow\:{x}={z}\:……\left(\mathrm{1}\right) \\ $$$${Also}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{{x}−{y}−{z}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:{x}−{y}−{z}=\mathrm{1}\:…….\left(\mathrm{2}\right)…
Question Number 45876 by rahul 19 last updated on 17/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18 $$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{29}}\: \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{c}}=\overset{\rightarrow} {{c}}×\overset{\rightarrow}…
Question Number 45856 by rahul 19 last updated on 17/Oct/18 $${Prove}\:{that}\:{the}\:{length}\:{of}\:{the}\:{perpendicular} \\ $$$${from}\:{the}\:{origin}\:{to}\:{the}\:{plane}\:{passing} \\ $$$${through}\:{point}\:\overset{\rightarrow} {{a}}\:{and}\:{containing}\:{the} \\ $$$${line}\:\overset{\rightarrow} {{r}}=\overset{\rightarrow} {{b}}+\lambda\overset{\rightarrow} {{c}}\:{is}\:\frac{\left[\overset{\rightarrow} {{a}}\:\:\overset{\rightarrow} {{b}}\:\:\overset{\rightarrow} {{c}}\:\right]}{\mid\overset{\rightarrow} {{b}}×\overset{\rightarrow}…
Question Number 45809 by rahul 19 last updated on 17/Oct/18 Commented by rahul 19 last updated on 17/Oct/18 $${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}.…