Answer

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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.

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 \[\overrightarrow {OA} - \overrightarrow {OP} \] .

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 \[\overrightarrow {OA} - \overrightarrow {OP} \] .

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