Question

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).

Answer 1

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.

Complete step-by-step answer:
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 $\overrightarrow {OA} - \overrightarrow {OP}$ .