Determine the vector equation of a line passing through the point P(3,2,-1) and with a direction vector perpendicular to the line r=(2,-3,4)+s(1,1,-2), seR

would the answer be r-(3,2,-1)+s(1,1,1)

Let's check:

r1=(3,2,-1)+s<1,1,1>
When s=0, r passes through the point (3,2,-1).
Now check the product of the two direction vectors:
<1,1,1> dot <1,1,-2>
=1+1-2
=0
So the two lines are perpendicular
and r1 satisfies the required conditions.

Note: multiple lines can be perpendicular r, for example:
r2=(3,2,-1)+t<5,3,4>
would also work, since
<5,3,4>dot<1,1,-2>=0 as well.

To determine the vector equation of a line passing through the point P(3,2,-1) and with a direction vector perpendicular to the line r=(2,-3,4)+s(1,1,-2), we need to find a vector that is perpendicular to the direction vector of the line.

The direction vector of the line is given as (1, 1, -2).

To find a perpendicular vector, we can take the cross product of this direction vector with any other vector.

Let's take the cross product of the direction vector (1, 1, -2) with the vector (1, 0, 0):

(1, 1, -2) x (1, 0, 0) = ((1*-2)-(0*1), (-2*1)-(1*0), (1*0)-(1*-2))
= (-2, -2, 0)

Now we have a vector (-2, -2, 0) that is perpendicular to the line.

The vector equation of the line passing through point P(3, 2, -1) is:

r = P + tN

where r is a position vector on the line, P is the position vector of the point P, t is a scalar parameter, and N is the direction vector perpendicular to the line.

Substituting the values:

r = (3, 2, -1) + t(-2, -2, 0)

Therefore, the vector equation of the line passing through the point P(3,2,-1) and with a direction vector perpendicular to the line r=(2,-3,4)+s(1,1,-2) is:

r = (3, 2, -1) + t(-2, -2, 0)

To determine the vector equation of a line passing through the point P(3,2,-1), we need to find a direction vector for the line.

Given the line equation r = (2,-3,4) + s(1,1,-2), we can see that the direction vector of this line is (1,1,-2), which is the coefficient of s in the equation.

To find a direction vector perpendicular to (1,1,-2), we can take the cross product of this vector with any other vector. Let's choose the vector (1,0,0).

The cross product of (1,1,-2) and (1,0,0) can be found using the following formula:
(Ay * Bz - Az * By, Az * Bx - Ax * Bz, Ax * By - Ay * Bx)

Substituting the values:
(1 * 0 - (-2) * 0, (-2) * 1 - 1 * 0, 1 * 0 - 1 * 0) = (0, -2, 0)

Now, we have a direction vector perpendicular to (1,1,-2), which is (0, -2, 0).

Using this direction vector and the given point P(3,2,-1), we can write the vector equation of the line as:
r = (3,2,-1) + t(0,-2,0), where t is a scalar parameter.

Therefore, the vector equation of the line passing through the point P(3,2,-1) and with a direction vector perpendicular to the line r = (2,-3,4) + s(1,1,-2) is r = (3,2,-1) + t(0,-2,0).