
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
594.6k+ views
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 \[\overrightarrow {OA} - \overrightarrow {OP} \] .
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

How many 5 digit telephone numbers can be constructed class 11 maths CBSE

Draw a well labelled diagram of reflex arc and explain class 11 biology CBSE

What is the difference between noise and music Can class 11 physics CBSE

Trending doubts
In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE

Correct the following 1m1000cm class 11 physics CBSE

