ABCD-EFGH-is-cube-X-is-midpoint-EF-if-AB-6-cm-how-distance-AX-to-BD-

Question Number 2499 by Syaka last updated on 21/Nov/15

A B C D . E F G H i s c u b e, X i s m i d p o i n t E F

i f A B = 6 c m, h o w d i s t a n c e A X t o B D ? ?

Commented by prakash jain last updated on 21/Nov/15

What is AM ? Is X = M ?

Commented by Syaka last updated on 21/Nov/15

s o r r y S i r, I m i s t a k e w h e n w r i t e t h i s q u e s t i o n

a n d n o w I^{'} v e c o r r e c t e d

Answered by prakash jain last updated on 21/Nov/15

Let A (0, 0, 0)

A = (0, 0, 0) B = (6, 0, 0) C = (6, 6, 0) D = (0, 6, 0)

E = (0, 0, 6) F = (6, 0, 6) G = (6, 6, 6) H = (0, 6, 6)

X = (3, 0, 6)

direction vector {\vec{d}}_{1} = \vec{BD} = (- 6, 6, 0)

BD : \frac{x - 6}{- 6} = \frac{y}{6} = t, z = 0

x = 6 - 6 t, y = 6 t, z = 0 (some p o i n t Q on line BD)

direction vector {\vec{d}}_{2} = \vec{AX} = (3, 0, 6)

AX : \frac{x}{3} = \frac{z}{6} = s, y = 0

x = 3 s, z = 6 s, y = 0 (some point R on line AX)

\vec{QR} = (3 s - 6 + 6 t, - 6 t, 6 s)

To calculate the distance we need to solve

for s and t such \vec{QR} ⊥ \vec{BD} and \vec{QR} ⊥ \vec{AX} .

\vec{QR} ∙ \vec{BD} = 0 and \vec{QR} ∙ \vec{AX} = 0

Once s and t are found we can find coordinates

of Q and R and calculate the distance .

\vec{QR} ∙ \vec{BD} = 0 \Rightarrow - 18 s + 36 - 36 t - 36 t = 0

s + 4 t = 2 \dots . (1)

\vec{QR} ∙ \vec{AX} = 0 \Rightarrow 9 s - 18 + 18 t + 36 s = 0

5 s + 2 t = 2 \dots . (2)

From (1) and (2) t = \frac{4}{9}, s = \frac{2}{9}

x = 6 - \frac{24}{9} = \frac{10}{3}, y = \frac{8}{3}, z = 0 (p o i n t Q on line BD)

x = \frac{2}{3}, z = \frac{4}{3}, y = 0 (point R on line AX)

\vec{QR} is ⊥ to both BD and AX and hence its

length gives the distance .

QR = \sqrt{{(\frac{8}{3})}^{2} + {(\frac{8}{3})}^{2} + {(\frac{4}{3})}^{2}} = \frac{1}{3} \sqrt{128 + 16} = 4