Question Video: Finding the Equation of a Plane in Vector Form | Nagwa Question Video: Finding the Equation of a Plane in Vector Form | Nagwa

# Question Video: Finding the Equation of a Plane in Vector Form Mathematics • Third Year of Secondary School

## Join Nagwa Classes

Find the vector form of the equation of the plane containing the two straight lines π«β = (π’ β π£ β 3π€) + π‘β(3π’ + 3π£ + 4π€) and π«β = (βπ’ β 2π£ β 3π€) + π‘β(βπ’ β 2π£ β 4π€).

04:11

### Video Transcript

Find the vector form of the equation of the plane containing the two straight lines π« sub one is equal to π’ minus π£ minus three π€ plus π‘ sub one multiplied by three π’ plus three π£ plus four π€ and π« sub two is equal to negative π’ minus two π£ minus three π€ plus π‘ sub two times negative π’ minus two π£ minus four π€.

In this question, weβre asked to find the vector form of the equation of a plane. And to do this, weβre given the equation of two straight lines contained in the plane. So to answer this question, letβs start by recalling the vector form of the equation of a plane. Itβs an equation of the form the dot product between vectors π§ and π« is equal to π, where π§ is a normal vector to the plane and π is a constant.

So to find the vector form of the equation of a plane, we need to find a normal vector to the plane. And we can do this by noting weβre given two lines contained within the plane. And this means that both lines run parallel to the plane. So the direction vectors of both lines are parallel to the plane. And we can see the direction vectors of both lines. The direction vector of the first line, π sub one, is three π’ plus three π£ plus four π€. And the direction vector of the second line, π sub two, is negative π’ minus two π£ minus four π€.

Therefore, for a vector to be normal to the plane, it must be orthogonal to both direction vectors of the lines. And in particular, we can find a normal vector to the direction vectors by calculating their cross product. And we do this by calculating the determinant of the three-by-three matrix where the first row are the unit directional vectors π’, π£, and π€ and the second and third row of this matrix are the components of the vectors. So our normal vector π§ is the determinant of the three-by-three matrix π’, π£, π€, negative one, negative two, negative four, three, three, four. And itβs worth noting we can switch the order of our second and third row. The only difference is weβll multiply our normal vector by negative one.

We can now evaluate this determinant by expanding over the first row. We get π’ times the determinant of the matrix negative two, negative four, three, four minus π£ multiplied by the determinant of the matrix negative one, negative four, three, four plus π€ times the determinant of the matrix negative one, negative two, three, three. Now, all we need to do is evaluate the determinant of these three matrices. We can do this by recalling the determinant of a two-by-two matrix is the difference in the product of its diagonals. Evaluating the determinant of each matrix and simplifying, we get π’ multiplied by negative eight plus 12 minus π£ times negative four plus 12 plus π€ times negative three plus six.

Now, all we need to do is evaluate the components of this vector. We get four π’ minus eight π£ plus three π€. So we found our normal vector to the plane π§. All we need to do now is find the constant value of π. And we can do this by substituting our normal vector π§ into the equation of the plane and using a known point on the plane.

Remember, the vector π« is the position vector of any point on the plane. So we need to find the position vector of a point on the plane. And we can do this by noting both straight lines are contained within the plane. And in particular, we can choose the vector π« sub zero equal to π’ minus π£ minus three π€ to be the position vector of a point on the plane, since it lies on the first straight line. So our value of the constant π will be the dot product between the normal vector and this position vector of the point on the line.

So all we need to do is calculate the dot product between the vector four, negative eight, three and the vector one, negative one, negative three. We can do this by recalling to find the dot product of two vectors of equal dimension, all we need to do is find the sum of the products of the corresponding components. Evaluating this then gives us four times one plus negative eight multiplied by negative one plus three times negative three, which we can then evaluate. We get four plus eight minus nine, which is equal to three, which shows that our value of π is three.

So we can substitute our expression for the normal vector π§ and the constant π into the vector form of the equation of the plane, which then gives us that the dot product between the vector four, negative eight, three and π« is equal to three is the vector form of the equation of the plane containing the two straight lines π« sub one and π« sub two.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions