Solving a Simple Vector Problem

[Einstein]

"simple" vector problem

I need help on the second part of this question:

The points A and B have position vectors (4i -11j +4k) and (7i +j +7k) respectively.

a) Find vector equation of the line passing through the two points in terms of a parameter t.

My answer: r = 4i -11j +4k + t(3i +12j +k)

b) Find the position vector of the point P on AB such that OP is perpendicular to AB where O is the origin.

I have attempted this many times but I do not know how to start. Can anyone please tell me how to. I am on the level capable of understanding scalar and dot product and converting vector equations into cartesian equations so i appreciate if you can keep it as simple as this (not too complex) and understandble. Thanks. By the way the answer to b) is 6i -3j +6k

 

[arcnets]

Originally posted by Einstein
My answer: r = 4i -11j +4k + t(3i +12j +k)

Almost correct. Should read
r = 4i -11j +4k + t(3i +12j +3k)

For b), just remember that two vectors are perpendicular (orthogonal) WRT to each other, if their scalar product is zero.

 

[M98Ranger]

To solve part b) of this problem, we can use the dot product to find the position vector of point P. Since we know that OP is perpendicular to AB, we can use the fact that the dot product of two perpendicular vectors is equal to 0.

First, let's find the vector AB by subtracting the position vector of point A from the position vector of point B:

AB = (7i +j +7k) - (4i -11j +4k)
= 3i +12j +3k

Next, we need to find the unit vector of AB by dividing the vector AB by its magnitude:

|AB| = √(3^2 + 12^2 + 3^2) = √(144) = 12

uAB = (3i +12j +3k)/12 = (1/4)i + (1/4)j + (1/4)k

Now, we can use the dot product to find the position vector of point P. We know that the dot product of OP and AB is equal to 0, so we can set up the following equation:

OP · AB = 0

Substituting the values we have found, we get:

(x,y,z) · (3i +12j +3k) = 0

Using the distributive property, we can expand this to:

3x +12y +3z = 0

We also know that the position vector of point P must lie on the line AB, so we can set up the following equation using the vector equation we found in part a):

(x,y,z) = (4i -11j +4k) + t(3i +12j +3k)

Substituting this into the equation we found earlier, we get:

3(4i -11j +4k) +12(-11j +4k) +3(4k) + t(3i +12j +3k) = 0

Expanding this equation, we get:

12i -33j +12k -132j +48k +12k + 3ti +12tj +3tk = 0

Simplifying, we get:

(3t+12)i + (-33+12t)j + (12+3t)k = 0

Since the coefficients of i, j