If xy− plane and yz− plane divides the line segment joining A(2,4,5) and B(3,5,−4) in the ratio a:b and p:q respectively then value of may be

Engineering

Mathematics

Distance from a Plane

Equation of Straight Line in 3D and Distance Of a Point From Line

Question

If xy− plane and yz− plane divides the line segment joining A(2,4,5) and B(3,5,−4) in the ratio a:b and p:q respectively then value of $(\frac{a}{b} + \frac{p}{q})$ may be

$\frac{7}{5}$

$\frac{23}{12}$

$\frac{7}{12}$

$\frac{21}{10}$

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Let the point on xy-plane which divide line AB in the ratio a:b be (x₁,y₁,0) and the point on yz-plane which divide line AB in the ratio p:q be (0,y₂,z₂).

Then

(x_{1}, y_{1}, 0) = (\frac{3 a + 2 b}{a + b}, \frac{5 a + 4 b}{a + b}, \frac{- 4 a + 5 b}{a + b})

⇒ \frac{- 4 a + 5 b}{a + b} = 0

⇒ 4a = 5b

\Rightarrow \frac{a}{b} = \frac{5}{4}

and

(0, y_{2}, z_{2}) = (\frac{3 p + 2 q}{p + q}, \frac{5 p + 4 q}{p + q}, \frac{- 4 p + 5 q}{p + q})

\Rightarrow \frac{3 p + 2 q}{p + q} = 0

⇒ 3p = –2q

\Rightarrow \frac{p}{q} = \frac{- 2}{3}

Now,

\frac{a}{b} + \frac{p}{q} = \frac{5}{4} - \frac{2}{3} = \frac{15 - 8}{12} = \frac{7}{12}