The position vectors a,b,c of the points A,B,C relative to the origin O are ai, bj and ck respectively.Find:
1) a vector equation for the line l throught O which is normal to the plane ABC.
2) a vector equation of the plane ABC
The answers should be:
1) r=t(bc i +ac j +ab k) and
2)r.(bc i +ac j +ab k) = abc
Can someone please explain how to work this question? Thanks!
To find the vector equation for the line l through O that is normal to the plane ABC, we need to find the normal vector of the plane. The normal vector is perpendicular to any vector lying in the plane. In this case, we can find the normal vector by taking the cross product of two vectors lying in the plane.
Let's find vectors AB and AC:
AB = B - A = (bj - ai)
AC = C - A = (ck - ai)
Now, we can find the normal vector N by taking the cross product of AB and AC:
N = AB x AC = (bj - ai) x (ck - ai)
To simplify the expression, expand the cross product:
N = [(bj x ck) - (ai x ck) - (bj x ai)]
Now, let's find the vector equation for the line l:
Since the line passes through the origin O, the position vector for any point P on the line can be written as r = t(OP), where t is a scalar.
We want the line to be normal to the plane, so the direction vector of the line should be the normal vector N.
Therefore, the vector equation for the line l is:
r = tN = t[(bj x ck) - (ai x ck) - (bj x ai)]
Now, let's move on to finding the vector equation of the plane ABC.
A vector equation of a plane can be written in the form:
r . N = d
Where r is the position vector of any point on the plane, N is the normal vector of the plane, and d is a constant.
In our case, we already have the normal vector N from the previous step. To find the constant d, we can substitute the coordinates of any point in the plane (e.g., A) into the equation and solve for d.
Using point A (ai) and the normal vector N, the equation becomes:
(ai) . N = d
Expanding the dot product:
(ai) . [(bj x ck) - (ai x ck) - (bj x ai)] = d
Simplifying further:
(ai) . (bj x ck) - (ai) . (ai x ck) - (ai) . (bj x ai) = d
Now, recall the properties of the scalar triple product: a . (b x c) = b . (c x a) = c . (a x b).
Using this property, we can simplify the equation to:
(ai) . (bj x ck) - (ai) . (ck x ai) = d
Expanding the dot products:
(ai) . (bj x ck) - [(ai x ai) . ck] = d
Since the cross product of a vector with itself is zero, (ai x ai) = 0:
(ai) . (bj x ck) = d
Therefore, the vector equation of the plane ABC is:
r . N = d
r . [(bj x ck) - (ai x ck) - (bj x ai)] = d
Simplifying further:
r . (bj x ck) - r . (ai x ck) - r . (bj x ai) = d
Finally, we get the vector equation of the plane ABC as:
r . (bj x ck) - r . (ai x ck) - r . (bj x ai) = abc
Sure! I can explain how to work through this question.
To find a vector equation for the line l through the origin O that is normal to the plane ABC, we need to find a vector that is perpendicular to the plane. Let's call this vector N.
1) Finding the vector N:
To find N, we can take the cross product of the vectors AB and AC. The cross product of two vectors gives us a vector that is perpendicular to both of the original vectors.
N = AB x AC
To find the cross product, we use the determinant method. We can set up a 3x3 matrix as follows:
| i j k |
| ai bj ck |
| ai bj ck |
Now, we expand the determinant along the top row:
N = (bj*ck - cj*bk)i - (ai*ck - ak*ci)j + (ai*bk - ak*bj)k
Simplifying further, we have:
N = (bc i + ac j + ab k)
So the vector equation for the line l through O that is normal to the plane ABC is:
r = t(bc i + ac j + ab k)
2) Finding the vector equation of the plane ABC:
To find the vector equation of the plane ABC, we can use the dot product of the normal vector N and the position vector of any point on the plane (let's say the origin O). The dot product of two vectors gives us a scalar outcome.
N · O = abc
Substituting the values of N and O, we have:
(bc i + ac j + ab k) · (0 i + 0 j + 0 k ) = abc
Simplifying, we have:
r · (bc i + ac j + ab k) = abc
Therefore, the vector equation of the plane ABC is:
r · (bc i + ac j + ab k) = abc