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## Homework Statement

Find the vector equation of the line through 4,5,5 that meets (x-11)/3=7+8=z+4 at right angles

## Homework Equations

## The Attempt at a Solution

i have no idea where to start.

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- Thread starter choob
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- #1

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Find the vector equation of the line through 4,5,5 that meets (x-11)/3=7+8=z+4 at right angles

i have no idea where to start.

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Dick

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- #3

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find a vector equ ation of the line through the point (4,5,5) that meets the line (x-11)/3=(y+8)/1=(z-4)/1 at right angles

i understand that the dot product of the vector i have to find and of the line given will be equal to zero. other than that, im completely lost.

- #4

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4-(11+3t), 5-(-8+t), 5-(4+t) (starting point of the vector subtract a point on the line)

then using dot product equation, i isolated t to -7/11

is there anything flawed in this logic?

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Dick

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[tex]\frac{x-11}{3}=\frac{y+8}{1}=\frac{z-4}{1} [/tex]

directly gives you the direction ratios of the required line, isnt it? What would they be?

- #7

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there are an infinite number of possible normals to the original line

- #8

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[tex]

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} [/tex]

where, (a,b,c) are the direction ratios and [tex](x_1,y_1,z_1)[/tex] is the point through which the line passes!

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