Show that the points -3 k- and 3-k- are equidistant from the plane r -5 -2 -7 k-9-0 and lies on opposite side of it-

To show that these given points î ĵ+3k̂ and 3(î+ĵ+k̂) are equidistant from the plane r⃗.(5î+2ĵ 7k̂)+9=0,we first find out the mid point of the points which is ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Plane

Show that the...

Question

Show that the points $^i -^j + 3^k$ and $3 (^i +^j +^k)$ are equidistant from the plane $\to r . (5^i + 2^j - 7^k) + 9 = 0$ and lies on opposite side of it.

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Solution

To show that these given points $^i -^j + 3^k$ and $3 (^i +^j +^k)$ are equidistant from the plane $\to r . (5^i + 2^j - 7^k) + 9 = 0$ ,we first find out the mid-point of the points which is $2^i +^j + 3^k$ .

On substituting $\to r$ by the mid-point in plane,we get

LHS= $(2^i +^j + 3^k) . (5^i + 2^j - 7^k) + 9$

=10+2-21+9=0

=RHS

Hence,the two points lie on opposite sides of the plane are equidistant from the plane.

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Q. The position vectors of points A and B are

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Show that the points ^i−^j+3^k and 3(^i+^j+^k) are equidistant from the plane →r.(5^i+2^j−7^k)+9=0 and lies on opposite side of it.

Show that the points $^i -^j + 3^k$ and $3 (^i +^j +^k)$ are equidistant from the plane $\to r . (5^i + 2^j - 7^k) + 9 = 0$ and lies on opposite side of it.