Consider line L, given below. x = 9 + t, y = 7 + t, z = 5 + 4t, t is in ℝ (a) Find point P that belongs to the line and direction vector v of the line. Express v in component form. P = Correct: Your answer is correct. v = Correct: Your answer is correct. (b) Find the distance from the origin to line L.

Answer :

Answer:

See the explanation.

Step-by-step explanation:

(a)

The equation of the line, L is given by x = 9 + t, y = 7 + t, z = 5 + 4t; where t is a real number.

In the above equation, if we put t = 0, then x = 9, y = 7 and z = 5.

Hence, the point (9, 7, 5) lies on the given line. We can say it as P.

The direction vector, v is {1, 1, 4}.

(b)

The equation of the given line can also be written as [tex]x-9 = y - 7 = \frac{z-5}{4}[/tex].

The vector, OP = (9, 7, 5), where O is the origin.

The direction vector v = {1, 1, 4}.

The cross product of OP and v is [tex]\left[\begin{array}{ccc}i&j&k\\9&7&5\\1&1&4\end{array}\right][/tex] = {23, -31, 2}.

Hence, the distance is [tex]\frac{\sqrt{(23)^2 + (-31)^2 + 2^2} }{\sqrt{1^ + 1^2 + 4^2} } = \frac{\sqrt{1494} }{\sqrt{18}} = \sqrt{83}[/tex] ≅9.

Other Questions