Question

From a point Q (a, b, c) perpendiculars QA and QB are drawn to the YZ and ZX planes respectively. What is the vector equation of the plane OAB?

Answer 1

Hint: Find the coordinates of the points A and B and then, find the vectors OA and OB. Take the cross product of these vectors to find the normal vector to the plane. Represent the plane in its vector equation form.

Complete step by step answer:
The coordinates of the point Q is given as (a, b, c).

It is given that QA is perpendicular to the YZ plane and A lies of the YZ plane. We know that the x coordinate of the point on the YZ plane is zero. Hence, the coordinates of the point A is (0, b, c).

It is given that QB is perpendicular to the ZX plane and B lies of the ZX plane. We know that the y coordinate of the point on the ZX plane is zero. Hence, the coordinates of the point B is (a, 0, c).

The vectors OA and OB are as follows:

\[\overrightarrow {OA} = b\hat j + c\hat k\]

\[\overrightarrow {OB} = a\hat i + c\hat k\]

The normal to the plane OAB is along the cross product of vectors OA and OB.

The cross product of the vectors OA and OB is given as follows:

\[\overrightarrow n = \overrightarrow {OA} \times \overrightarrow {OB} \]

\[\overrightarrow n = \left| {\begin{array}{*{20}{c}}

  {\hat i}&{\hat j}&{\hat k} \\

  0&b&c \\

  a&0&c

\end{array}} \right|\]

\[\overrightarrow n = bc\hat i + ac\hat j - ab\hat k...........(1)\]

The vector equation of a plane is given as:

\[\vec r.\vec n = 0\]

Hence, using equation (1), the vector equation of the plane OAB is given as follows:

\[\vec r.(bc\hat i + ac\hat j - ab\hat k) = 0\]

Hence, the vector equation of the plane OAB is \[\vec r.(bc\hat i + ac\hat j - ab\hat k) = 0\].


Note: You can also find the equation of the plane using a three point formula and then convert it into a vector equation by taking the coefficients as components of the normal vector.