Question

Let the vectors \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \]such that \[\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 4\,{\rm{and}}\,\left| {\overrightarrow c } \right| = 4\]. If the projection of vector b on vector a is equal to the projection of vector c on vector a and b is perpendicular to vector c, then find the value of \[\left| {\overrightarrow a + \overrightarrow b - \overrightarrow c } \right|\].

Answer 1

Hint: First establish a relation between a, b and c by the given information, that is the projection of vector b on vector a is equal to the projection of vector c on vector a, then form an equation by the information that, b is perpendicular to the vector c.
Now, calculate and take the square root of both sides to obtain the required value.

Formula used:
The projection formula of vector P on vector Q is \[\dfrac{{\overrightarrow P \cdot \overrightarrow Q }}{{\left| {\overrightarrow Q } \right|}}\].

Complete step by step solution:
It is given that \[\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 4\,{\rm{and}}\,\left| {\overrightarrow c } \right| = 4\].
Now, projection of vector b on vector a is \[\dfrac{{\overrightarrow b \cdot \overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \dfrac{{\overrightarrow b \cdot \overrightarrow a }}{2}\] and projection of vector c on vector a is \[\dfrac{{\overrightarrow c \cdot \overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \dfrac{{\overrightarrow c \cdot \overrightarrow a }}{2}\].
It is given that \[\dfrac{{\overrightarrow b \cdot \overrightarrow a }}{2} = \dfrac{{\overrightarrow c \cdot \overrightarrow a }}{2}\]
That is, \[\overrightarrow b \cdot \overrightarrow a = \overrightarrow c \cdot \overrightarrow a \]
And as b vector is perpendicular to c vector therefore, \[\overrightarrow b \cdot \overrightarrow c = 0\]
Now,
 \[{\left(| { \overrightarrow a + \overrightarrow b - \overrightarrow c}| \right)^2}={|\overrightarrow a|^{2}} + {|\overrightarrow b|^{2}} + {|\overrightarrow c|^{2}} + 2\cdot \overrightarrow a\cdot \overrightarrow b – 2\cdot \overrightarrow b\cdot \overrightarrow c – 2\cdot \overrightarrow c\cdot \overrightarrow a\]
\[ = {2^2} + {4^2} + {4^2} + 2\overrightarrow a \cdot \overrightarrow b - 2 \times 0 - 2\overrightarrow a \cdot \overrightarrow b \]
 \[ = 4 + 16 + 16\]
 \[ = 36\]
Therefore, \[\left| {\overrightarrow a + \overrightarrow b - \overrightarrow c } \right|\]=6
Hence the required value is 6.

Note: The formula of \[{\left( {a + b - c} \right)^2}\] is \[{a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ca\], sometime students get confused with the formula.