The points with position vectors\[20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge \],\[5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge \] and \[10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge \] are collinear. The value of p is:
Answer
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Hint: If we have two vectors A and B then \[\overrightarrow {AB} \] is given as \[\overrightarrow B - \overrightarrow A \]. Two vectors A and B to be collinear, angle between the vector A and B made by the given position vectors should be 0 or 180 degree.
Complete step-by-step answer:
Suppose position vector\[\overrightarrow A = 20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge \],\[\overrightarrow B = 5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge \] and \[\overrightarrow C = 10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge \].
Now, \[\overrightarrow {AB} = \overrightarrow B - \overrightarrow A \]
\[
= 5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge - (20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge ) \\
= - 15\mathop i\limits^ \wedge - (1 + p)\mathop j\limits^ \wedge \\
\]
Similarly, \[\overrightarrow { = BC} = \overrightarrow C - \overrightarrow B \]
\[
= 10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge - (5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge ) \\
= 5\mathop i\limits^ \wedge - 12\mathop j\limits^ \wedge \\
\]
As we know they are collinear, the angle between the vector AB and BC must be 0 or 180 degree.
\[
\Rightarrow \overrightarrow {AB} \times \overrightarrow {BC} = 0 \\
\Rightarrow \left( { - 15\mathop i\limits^ \wedge - (1 + p)\mathop j\limits^ \wedge } \right) \times \left( {5\mathop i\limits^ \wedge - 12\mathop j\limits^ \wedge } \right) = 0 \\
\Rightarrow 0\mathop i\limits^ \wedge + ( - 15)( - 12) + (1 + p)(5) + 0\mathop j\limits^ \wedge = 0 \\
\Rightarrow 180 + 5 + 5p = 0 \\
\Rightarrow 5p = - 185 \\
\Rightarrow p = - 37 \\
\]
Required value of p is -37.
Note: Collinear points are the points that lie on a single line and therefore the angle between them will either be 0 or 180 degree. So, if two vectors \[\overrightarrow A \] and \[\overrightarrow B \]are collinear then we can write it as \[\overrightarrow A = n\overrightarrow B \].
Complete step-by-step answer:
Suppose position vector\[\overrightarrow A = 20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge \],\[\overrightarrow B = 5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge \] and \[\overrightarrow C = 10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge \].
Now, \[\overrightarrow {AB} = \overrightarrow B - \overrightarrow A \]
\[
= 5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge - (20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge ) \\
= - 15\mathop i\limits^ \wedge - (1 + p)\mathop j\limits^ \wedge \\
\]
Similarly, \[\overrightarrow { = BC} = \overrightarrow C - \overrightarrow B \]
\[
= 10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge - (5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge ) \\
= 5\mathop i\limits^ \wedge - 12\mathop j\limits^ \wedge \\
\]
As we know they are collinear, the angle between the vector AB and BC must be 0 or 180 degree.
\[
\Rightarrow \overrightarrow {AB} \times \overrightarrow {BC} = 0 \\
\Rightarrow \left( { - 15\mathop i\limits^ \wedge - (1 + p)\mathop j\limits^ \wedge } \right) \times \left( {5\mathop i\limits^ \wedge - 12\mathop j\limits^ \wedge } \right) = 0 \\
\Rightarrow 0\mathop i\limits^ \wedge + ( - 15)( - 12) + (1 + p)(5) + 0\mathop j\limits^ \wedge = 0 \\
\Rightarrow 180 + 5 + 5p = 0 \\
\Rightarrow 5p = - 185 \\
\Rightarrow p = - 37 \\
\]
Required value of p is -37.
Note: Collinear points are the points that lie on a single line and therefore the angle between them will either be 0 or 180 degree. So, if two vectors \[\overrightarrow A \] and \[\overrightarrow B \]are collinear then we can write it as \[\overrightarrow A = n\overrightarrow B \].
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