Write the sum of intercepts cut off by the plane r -2 -k-5-0 on the three axes-

Firstly we find the cartesian equation of the given plane. nbsp; r. (2î+ĵ k̂) 5 = 0 (xî+yĵ+zk̂). (2î+ĵ k̂)=0 ⇒ 2x+y z = 5 is the required cartesian equatio ...

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Standard XII

Mathematics

Family of Planes Passing through the Intersection of Two Planes

Write the sum...

Question

Write the sum of intercepts cut off by the plane $\to r . (2^i +^j -^k) - 5 = 0$ on the three axes.

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Solution

Firstly we find the cartesian equation of the given plane.

$\to r . (2^i +^j -^k) - 5 = 0$

$(x^i + y^j + z^k) . (2^i +^j -^k) = 0$

$\Rightarrow 2 x + y - z = 5$ is the required cartesian equation.

$\Rightarrow \frac{2 x}{5} + \frac{y}{5} - \frac{z}{5} = 1$

$\Rightarrow \frac{x}{5 / 2} + \frac{y}{5} + \frac{z}{- 5} = 1$

Comparing this with $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , we have $a = \frac{5}{2}, b = 5$ and $c = - 5$

Thus, the given plane cuts off intercepts $\frac{5}{2}, 5, - 5$ along the three axes.

Hence, required sum of intercepts is

= $\frac{5}{2} + 5 - 5$

= $\frac{5}{2}$

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Write the sum of intercepts cut off by the plane →r.(2^i+^j−^k)−5=0 on the three axes.

Write the sum of intercepts cut off by the plane $\to r . (2^i +^j -^k) - 5 = 0$ on the three axes.