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How to find the distance using dot product from point $A$ to the line through $B$ and $C$?

The points are $A = (2, -6, 1)$, $B = (3, 4, -2)$ and $C = (7, -1, 5)$.

I have tried this method: $$d=\frac{||BC \times BA||}{||BC||}$$

Please tell me how to find the distance using dot product.

Smi
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Alee
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2 Answers2

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The line throguh points $\;B,C\;$ can be put (parametrized) as

$$l_{BC}: B+t\vec{BC}=(3,4,-2)+t(4,-5,7)=(3+4t\,,\,4-5t\,,\,-2+7t)\;,\;\;t\in\Bbb R$$

Thus, a vector from $\;A\;$ to a general point $\;Q\;$ on the line is parametrized as

$$\vec{AQ}=(3+4t,\,4-5t,\,-2+7t)-(2,-6,1)=(1+4t\,,\,10-5t\,,\,-3+7t)$$

Finally, we need that

$$\;0=\langle A\,,\,\vec{AQ}\rangle=2+8t+12+30t-3+7t\implies45t=-11\implies t=-\frac{45}{11}$$

Complete now the exercise.

DonAntonio
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  • Your $\vec{AQ}$ is incorrect. – linear_combinatori_probabi Sep 25 '20 at 04:02
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    @linear_combinatori_probabi Indeed, though merely writing "it is incorrect" without saying what the mistake is isn't as helpful. Anyway,. this is more than 6.5 years old and the OP, or whoever is trying to learn somethiong from this, should check the work carefully.. – DonAntonio Sep 25 '20 at 06:14
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HINT:
use the property that v1.v2=0 , if v1,v2 are perpendicular.

OK.so, let A(2,-6,1); B(3,4,-2); C(7,-1,5) be given. Now lets find BC vector, which comes BC(4,-5,7).
Now suppose a point P(x,y,z) in between B and C.and draw perpendicular from A to this point P .
We can find AP vector which is AP(x-2,4+6,z-1).
now dot product of AP and BC is zero. From here you get P.Once you have P. The |AP| is the required distance

ketan
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  • Plz make me more clear I didn't understand – Alee Mar 08 '14 at 10:33
  • @ketan, your answers wont't be fully appreciated (and perhaps not even read) if you don't learn how to use LaTeX to properly write mathematics in this site...a pity. – DonAntonio Mar 08 '14 at 11:19
  • Answer is not right by your method,answer is 7.75 but I'm getting 69 by your method,Is point P is(0,0,45/7)? – Alee Mar 08 '14 at 12:00