In the triangle with vertices A \left(\right. 1 , - 1 , 2 \left.\right) , B \left(\right. 5 , - 6 , 2 \left.\right) and C \left(\right. 1 , 3 , - 1 \left.\right) find the altitude n = \mid B D \mid.

Engineering

Mathematics

Line of Greatest Slope

Question

In the triangle with vertices

A (1, - 1, 2), B (5, - 6, 2)

and

C (1, 3, - 1)

find the altitude

n = ∣ B D ∣

5

10

5 \sqrt{2}

\frac{10}{\sqrt{2}}

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Given

A (1, - 1, 2), B (5, - 6, 2), C (1, 3, - 1)

Let the coordinates of

D

D (h, k, l)

the equation of line

A C

\frac{x - 1}{0} = \frac{y + 1}{4} = \frac{z - 2}{- 3}

So the point of line

A C

will be in the form of

(1, 4 t - 1, 2 - 3 t)

Since

D

lies on

A C

, we have

h = 1, k = 4 t - 1, l = 2 - 3 t

The drs of

B D

are

h - 5, k + 6, l - 2

and the drs of

A C

are

0, 4, - 3

Since

B D

and

A C

are perpendicular, we have

4 (k + 6) - 3 (l - 2) = 0

\Rightarrow 4 (4 t + 5) - 3 (- 3 t) = 0

\Rightarrow t = - \frac{4}{5}

So, coordinates of

D

are

D (1, - \frac{21}{5}, \frac{22}{5})

So, the value of

n = \sqrt{16 + \frac{81}{25} + \frac{144}{25},} = \sqrt{25} = 5

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