Elementary linear algebra questions

[tigerseye]

Could someone help me please?

1.) Find the scalar equation of the line containing P(2, -1, 3) and perpendicular to the lines [x y z]^T=[4 -1 2]^T + t[7 0 1] and [x y z]^T=[-2 0 1] + t[2 3 0]^T.

2.)Find all points C on the line through A(1, -1, 2) and B(2, 0, 1) such that vectors llACll= 2 llBCll.

 

[mathwonk]

as they say here: what have you tried?

 

[thepatientmental]

Sure, I can help you with these questions. Let's start with the first one.

1.) To find the scalar equation of the line perpendicular to the given lines, we first need to find the direction vector of the line. This can be done by taking the cross product of the direction vectors of the two given lines. So, the direction vector of the perpendicular line will be:

d = [7 0 1] x [2 3 0] = [3 -7 14]

Now, we can use the point-slope form of a line to find the equation. The point P(2, -1, 3) lies on the line, so we can use it as our point. The point-slope form is given by:

(x-x1)/a = (y-y1)/b = (z-z1)/c

where (a, b, c) is the direction vector and (x1, y1, z1) is the given point. Plugging in the values, we get:

(x-2)/3 = (y+1)/(-7) = (z-3)/14

This is the scalar equation of the line containing P(2, -1, 3) and perpendicular to the given lines.

2.) To find the points C on the line through A(1, -1, 2) and B(2, 0, 1) such that vectors llACll= 2 llBCll, we can use the distance formula between two points. The distance formula is given by:

d = √((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

where (x1, y1, z1) and (x2, y2, z2) are the two points. We can set up two equations using this formula, one for the distance between A and C and one for the distance between B and C. Since we want the distances to be equal, we can set the two equations equal to each other:

√((x-1)^2 + (y+1)^2 + (z-2)^2) = 2√((x-2)^2 + y^2 + (z-1)^2)

Simplifying and squaring both sides, we get:

(x-1)^2 + (y+1)^2 + (z-