Question

Find the vector and cartesian equation of the line passing through the point (213) and perpendicular to line x-1/1=y-2/2=z-3/3

Answer 1

Answer:

Step-by-step explanation:

Let the foot of the perpendicular from point A (2, 1, 3) to the line

x-1/1=y-2/2=z-3/3 be B

But Any point on the line x-1/1=y-2/2=z-3/3 will be of the form

(1 + k, 2 + 2k, 3 +3k).

So B is of the form (1 + k, 2 + 2k, 3 +3k).

Dr's of the line AB will be (1 - k, -1 -2k, -3k)

Dr's of the given line (1, 2, 3)

But if 2 lines are perpendicular, then the dot product of thier dr's

would be 0.

Hence,

(1-k) + 2(-1-2k) +3(-3k) = 0

=>1-k-2-4k-9k = 0

=>14k = -1

=> k = -1/14.

Hence Dr's of line AB are

(15/14, 12/14, 3/14) or

they are proportional to

(5, 4, 1) Since we can ignore the proportional constant in dr's.

Now we have dr's and point on line AB,

hence the vector equation of line AB is

r = a +λb

where a = 2i + 1j + 3k

b = 5i + 4j +k

So, r = (2i + j + 3k) + λ(5i + 4j +k) is the required equation in vector form.

Now to compute the equation of line in cartesian form let r = xi + yj + zk  Replace value of r and comparing each component, we get

x = 2 + 5λ, y = 1 + 4λ and z = 3 + λ

Finding λ from above 3, we get

x-2/5 = y-1/4 = z-3....Ans