The lengths of the intercepts on the co-ordinate axes made by the plane 5x + 2y + z – 13 = 0 are

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Question

This question was previously asked in

NDA (Held On: 16 Dec 2015) Maths Previous Year paper

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Answer 1

Concept:

The intercept form of the plane is given by ⇔ \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

Where ‘a’ is the x- intercept, ‘b’ is the y- intercept and ‘c’ is the z- intercept.

Calculation:

Given:

Equation of plane 5x + 2y + z – 13 = 0

⇒ 5x + 2y + z = 13

\(\Rightarrow \frac{{5{\rm{x\;}} + {\rm{\;}}2{\rm{y\;}} + {\rm{\;z\;}}}}{{13}} = 1\)

\(\Rightarrow \frac{x}{{\frac{{13}}{5}}} + \frac{y}{{\frac{{13}}{2}}} + \frac{z}{{\frac{{13}}{1}}} = 1\)

∴ Lengths of intercepts are \(\frac{{13}}{5},\frac{{13}}{2},\;13\) unit