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Step-by-Step Solution
Step 1: Understand the Given Information
We are given three vectors \overrightarrow{a} , \overrightarrow{b} , and \overrightarrow{c} with magnitudes:
\lvert \overrightarrow{a} \rvert = 2, \quad \lvert \overrightarrow{b} \rvert = 4, \quad \lvert \overrightarrow{c} \rvert = 4 .
Additionally, we have two important conditions:
The projection of \overrightarrow{b} on \overrightarrow{a} is equal to the projection of \overrightarrow{c} on \overrightarrow{a} .
\overrightarrow{b} is perpendicular to \overrightarrow{c} , which means \overrightarrow{b} \cdot \overrightarrow{c} = 0 .
We want to find the value of \left\lvert \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right\rvert .
Step 2: Apply the Projection Condition
The projection of \overrightarrow{b} on \overrightarrow{a} is given by
\dfrac{\overrightarrow{b} \cdot \overrightarrow{a}}{\lvert \overrightarrow{a} \rvert} .
Similarly, the projection of \overrightarrow{c} on \overrightarrow{a} is
\dfrac{\overrightarrow{c} \cdot \overrightarrow{a}}{\lvert \overrightarrow{a} \rvert} .
According to the given condition:
\frac{\overrightarrow{b} \cdot \overrightarrow{a}}{\lvert \overrightarrow{a} \rvert}
=
\frac{\overrightarrow{c} \cdot \overrightarrow{a}}{\lvert \overrightarrow{a} \rvert}.
Since \lvert \overrightarrow{a} \rvert \neq 0 , we can cancel \lvert \overrightarrow{a} \rvert on both sides to get
\overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a}.
Step 3: Use the Perpendicularity Condition
We are also given that \overrightarrow{b} \perp \overrightarrow{c} . This implies
\overrightarrow{b} \cdot \overrightarrow{c} = 0.
Step 4: Express \lvert \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \rvert in Terms of Dot Products
Let \left\lvert \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right\rvert = k.
Then,
k^2
=
\left( \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right)
\cdot
\left( \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right).
Expanding this dot product:
\[
k^2
=
\overrightarrow{a} \cdot \overrightarrow{a}
+
\overrightarrow{b} \cdot \overrightarrow{b}
+
\overrightarrow{c} \cdot \overrightarrow{c}
+
2\, \overrightarrow{a} \cdot \overrightarrow{b}
-
2\, \overrightarrow{b} \cdot \overrightarrow{c}
-
2\, \overrightarrow{a} \cdot \overrightarrow{c}.
\]
Recall:
\[
\lvert \overrightarrow{a} \rvert^2 = 2^2 = 4, \quad
\lvert \overrightarrow{b} \rvert^2 = 4^2 = 16, \quad
\lvert \overrightarrow{c} \rvert^2 = 4^2 = 16.
\]
Step 5: Simplify Using Given Conditions
From earlier conditions:
\overrightarrow{b} \cdot \overrightarrow{c} = 0.
\overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a}.
In the expansion of k^2 :
\[
k^2
=
(4) + (16) + (16)
+ 2 \left( \overrightarrow{a} \cdot \overrightarrow{b} \right)
- 2 \left( \overrightarrow{b} \cdot \overrightarrow{c} \right)
- 2 \left( \overrightarrow{a} \cdot \overrightarrow{c} \right).
\]
\overrightarrow{b} \cdot \overrightarrow{c} = 0 eliminates the -2(\overrightarrow{b} \cdot \overrightarrow{c}) term.
\overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a} implies 2(\overrightarrow{a} \cdot \overrightarrow{b}) - 2(\overrightarrow{a} \cdot \overrightarrow{c}) = 2(\overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{a} \cdot \overrightarrow{c}) = 0 .
Therefore,
\[
k^2
=
4 + 16 + 16
= 36.
\]
Step 6: Conclude the Magnitude
Since k^2 = 36 , we get
\[
k = \sqrt{36} = 6.
\]
Hence,
\[
\left\lvert \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} \right\rvert = 6.
\]
Final Answer
\boxed{6}
