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Question: Find the magnitude of the vector which starts at the point \(2\hat i + \hat j - 3\hat k\) and ends a...

Find the magnitude of the vector which starts at the point 2i^+j^3k^2\hat i + \hat j - 3\hat k and ends at 4i^j^k^4\hat i - \hat j - \hat k.

Explanation

Solution

Hint:Hint: In this question we will use one of the types of vectors i.e. position vector which is ab=ba\overrightarrow {ab} = \overrightarrow b - \overrightarrow a as it used to specify the positions of the vector. Where b\overrightarrow b is the ending point and a\overrightarrow a is the starting point of a vector .
Complete step-by-step solution -
According to the question two points where the vector starts and ends are given i.e. 2i^+j^3k^2\hat i + \hat j - 3\hat k and 4i^j^k^4\hat i - \hat j - \hat k respectively.
Now, let vector a=2i^+j^3k^\overrightarrow a = 2\hat i + \hat j - 3\hat k
b=4i^j^k^\overrightarrow b = 4\hat i - \hat j - \hat k
Use the Position Vector,
Hence ab=ba\overrightarrow {ab} = \overrightarrow b - \overrightarrow a
=4i^j^k^2j^j^+3k^ =2i^2j^+2k^  = 4\hat i - \hat j - \hat k - 2\hat j - \hat j + 3\hat k \\\ = 2\hat i - 2\hat j + 2\hat k \\\
Now, ab=(2)2+(2)2+(2)2\left| {\overrightarrow {ab} } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2}}
=4+4+4 =12 =23  = \sqrt {4 + 4 + 4} \\\ = \sqrt {12} \\\ = 2\sqrt 3 \\\
Hence the magnitude of the vector is 232\sqrt 3 .
NoteNote : In such types of questions where the magnitude of the vector has to find there we use the position vector i.e. if vector OA\overrightarrow {OA} is used to specify the position of a point AA relative to another point OO. This OA\overrightarrow {OA} is called the position vector of AA referred to OO as an origin. These concepts will help in solving vector questions so it is advisable to remember these concepts.