A  vector equation for the line through the point (-4,3,2) and perpendicular to the lines  L_2 = (-4,3,2) + (5,-5,-2)t and L_3 : (x,y,z) = (-4,3,2) + <-3,4,-3>t   is (x,y,z)= ???. Enter...

A  vector equation for the line through the point (-4,3,2) and perpendicular to the lines

 
L_2 = (-4,3,2) + (5,-5,-2)t

and

L_3 : (x,y,z) = (-4,3,2) + <-3,4,-3>t
 

is

(x,y,z)= ???.

Enter your answer in the form (x_0, y_0, z_0) + <a,b,c>t with t as your parameter.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the direction vectors of the given lines, such that:

`bar(L_2) = <5,-5,-2>`

`bar(L_3) = <-3,4,-3>`

You need to remember that evaluating the cross product of two vectors yields a vector that is perpendicular to both vectors, such that:

`bar(L_2) x bar(L_3) = [(bar i, bar j, bar k),(5,-5,-2),(-3,4,-3)]`

`bar(L_2) x bar(L_3) = 15 bar i + 20 bar k + 6 bar j - 15 bar k + 8 bar i + 15 bar j`

`bar(L_2) x bar(L_3) = 23 bar i + 21 bar j + 5 bar k`

Hence, evaluating the vector `bar n = bar(L_2) x bar(L_3)`  yields:

`bar n = <23,21,5>`

You may write the equation of the line passing through the point `(-4,3,2)` and perpendicular to the give lines, such that:

`L = (-4,3,2) + <23,21,5>*t`

Hence, evaluating the equation of the given line, under the given conditions, yields `L = (-4,3,2) + <23,21,5>*t.`

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