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Question: The points with position vectors\[20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge \],\[5\math...

The points with position vectors20i∧+pj∧20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge ,5i∧−j∧5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge and 10i∧−13j∧10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge are collinear. The value of p is:

Explanation

Solution

If we have two vectors A and B then AB→\overrightarrow {AB} is given as B→−A→\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 vectorA→=20i∧+pj∧\overrightarrow A = 20\mathop i\limits^ \wedge + p\mathop j\limits^ \wedge ,B→=5i∧−j∧\overrightarrow B = 5\mathop i\limits^ \wedge - \mathop j\limits^ \wedge and C→=10i∧−13j∧\overrightarrow C = 10\mathop i\limits^ \wedge - 13\mathop j\limits^ \wedge .
Now, AB→=B→−A→\overrightarrow {AB} = \overrightarrow B - \overrightarrow A

=5i∧−j∧−(20i∧+pj∧) =−15i∧−(1+p)j∧  = 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, =BC→=C→−B→\overrightarrow { = BC} = \overrightarrow C - \overrightarrow B

=10i∧−13j∧−(5i∧−j∧) =5i∧−12j∧  = 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.

⇒AB→×BC→=0 ⇒(−15i∧−(1+p)j∧)×(5i∧−12j∧)=0 ⇒0i∧+(−15)(−12)+(1+p)(5)+0j∧=0 ⇒180+5+5p=0 ⇒5p=−185 ⇒p=−37  \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 A→\overrightarrow A and B→\overrightarrow B are collinear then we can write it as A→=nB→\overrightarrow A = n\overrightarrow B .