Consider three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k. If p, q and r denote the position vector of three non-collinear points, then the equation of the plane containing these points is:

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This question was previously asked in

AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper

Answer 1

Given:

Three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k.

Concept:

The equation of plane passing through three points a , b and c is

\(\rm (\vec r-a)\cdot((b-a)\times(c-a))=0\)

Calculation:

Three vectors p = 2i + 3j + 4k, q = i + 4j - k and r = 2i + 3j + k.

Now,

q - p = - i + j - 5k

and r - p = 0i + 0j - 3k

Then the normal vector

\( \rm \vec n=(q-p)\times(r-p)=\begin{bmatrix}i&j&k\\-1&1&-5\\0&0&-3\end{bmatrix}\)

\(\rm \vec n = -3i-3j=-3(i+j)\)

Then the equation of plane is

\(\rm (\vec r-\vec p)⋅ \vec n=0\)

put all the values then

((xi + yj + zk) - (2i + 3j + 4k))⋅ (-3(i+j)) = 0

((xi + yj + zk) - (2i + 3j + 4k))⋅ (i+j) = 0

(xi + yj + zk) ⋅ (i+j) - (2i + 3j + 4k) ⋅ (i+j) = 0

x + y - 2 - 3 = 0

x + y - 5 = 0

Hence the option (2) is correct.

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