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# Using vectors show that the point A (-2, 3, 5), B (7, 0, -1), C (-3, -2, -5) and D (3, 4, 7) are such that AB and CD intersect at the point P (1, 2, 3). Verified
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Hint: Show that the point P lies on AB by showing that AB and AP are collinear. Then show that the point P lies on CD by showing that CP and CD are collinear. Since P lies on both AB and CD, it is the point of intersection of AB and CD.

From the given points we calculate the position vectors of each point from origin as follows:
$\overrightarrow {OA} = - 2i + 3j + 5k$
$\overrightarrow {OB} = 7i - 1k$
$\overrightarrow {OC} = - 3i - 2j - 5k$
$\overrightarrow {OD} = 3i + 4j + 7k$
$\overrightarrow {OP} = i + 2j + 3k$

We now find the vector $\overrightarrow {AP}$ as follows:
$\overrightarrow {AP} = \overrightarrow {OP} - \overrightarrow {OA}$
Substituting the vectors, we get:
$\overrightarrow {AP} = (i + 2j + 3k) - ( - 2i + 3j + 5k)$
Simplifying, we get:
$\overrightarrow {AP} = 3i - j - 2k........(1)$

Now, we find the vector $\overrightarrow {AB}$ as follows:
$\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA}$
Substituting the vectors, we get:
$\overrightarrow {AB} = (7i - 1k) - ( - 2i + 3j + 5k)$
Simplifying the expression, we get:
$\overrightarrow {AB} = 9i - 3j - 6k.........(2)$

Comparing equation (1) and equation (2), we observe:
$\overrightarrow {AB} = 3\overrightarrow {AP}$
Hence, the point P lies on the line AB.
We now find the vector $\overrightarrow {CP}$ as follows:
$\overrightarrow {CP} = \overrightarrow {OP} - \overrightarrow {OC}$
Substituting the vectors, we get:
$\overrightarrow {CP} = (i + 2j + 3k) - ( - 3i - 2j - 5k)$
Simplifying, we get:
$\overrightarrow {CP} = 4i + 4j + 8k........(3)$

Now, we find the vector $\overrightarrow {CD}$ as follows:
$\overrightarrow {CD} = \overrightarrow {OD} - \overrightarrow {OC}$
Substituting the vectors, we get:
$\overrightarrow {CD} = (3i + 4j + 7k) - ( - 3i - 2j - 5k)$
Simplifying the expression, we get:
$\overrightarrow {CD} = 6i + 6j + 12k.........(4)$
Comparing equation (3) and (4), we observe:
$\overrightarrow {CD} = \dfrac{3}{2}\overrightarrow {CP}$

Hence, the point P lies on the line CD.
Since, P lies on both the lines AB and CD, it is the point of intersection of the two lines.

Hence, we showed that AB and CD intersect at point P.

Note: The way we are asked to solve is clearly mentioned as using vectors, it is an error to solve using any other method other than vector method. Also, vector $\overrightarrow {AP}$ is $\overrightarrow {OP} - \overrightarrow {OA}$ and not $\overrightarrow {OA} - \overrightarrow {OP}$ .
Last updated date: 28th Sep 2023
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