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If from the point $P(a,b,c)$ perpendiculars $PA$ and $PB$ are drawn to $yz$ and $zx$ planes respectively then the vector equation of the plane $OAB$ is ?

$(a)\: \overrightarrow r.(a\hat i+b\hat j+c\hat k)=0 \:\qquad (b) \: \overrightarrow r.(\large\frac{\hat i}{a}+\frac{\hat j}{b}-\frac{\hat k}{c})=0$$ (c)\:\:\overrightarrow r.(a\hat i+b\hat j-c\hat k)=0\:\qquad (d)\:\:\overrightarrow r.(\large\frac{\hat i}{a}+\frac{\hat j}{b}+\frac{\hat k}{c})=0$

1 Answer

Given that $A$ is foot of $\perp$ from $P(a,b,c)$ on $yz$ plane.
$\therefore \;A=(0,b,c)$
Similarly it is given that $B$ is foot of $\perp$ from $P$ on $zx$ plane.
Norma vector $\overrightarrow n$ to the plane $OAB$ is $(0,b,c)\times (a,0,c)$
$i.e.,\:\overrightarrow n=\left |\begin {array}{ccc} \hat i &\hat j & \hat k \\0 & b & c \\ a & 0 & c\end {array}\right |=(bc,ac,-ab)$
Dividing by $abc$, we get $\overrightarrow n=(\large\frac{1}{a},\frac{1}{b},-\frac{1}{c})$
Since the plane passes through origin $O$, its vector eqn. is given by
$\overrightarrow r.(\large\frac{\hat i}{a}+\frac{\hat j}{b}-\frac{\hat k}{c})=0$
answered Jan 2, 2014 by rvidyagovindarajan_1

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