In the given figure, the angles $\angle BAQ = \angle CPQ = \angle CBQ = \frac{\pi}{2}$; and the lengths $QA = 3$ unit, $AB = 4$ unit, and $BC = 1$ unit. What is the length of $PQ$? }
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Using Cartesian coordinates is a highly efficient technique for solving complex geometric configurations with multiple right angles.
By setting one point as the origin, you can easily use vector slopes to determine the positions of all other vertices.
• Step 1 : Understanding the Question:
We are given a geometric figure with three right angles: \(\angle BAQ\)., \(\angle CPQ\)., and \(\angle CBQ\). are all \(\frac{\pi}{2}\).
The given lengths are \(QA = 3\) units, \(AB = 4\) units, and \(BC = 1\) unit.
We need to find the length of the segment \(PQ\).
• Step 2 : Key Formula or Approach:
The easiest and most rigorous way to solve this problem is to place the entire configuration on a Cartesian coordinate system.
Let \(Q\) be the origin \((0, 0)\).
Since \(\angle BAQ = \frac{\pi}{2}\)., we can place \(QA\) along the \(X\)-axis and \(AB\) as a vertical line.
We can then calculate the coordinates of the points \(B\)., \(C\)., and \(P\). using vectors.
• Step 3 : Detailed Explanation:
Let us establish the coordinates:
Let \(Q\) be at the origin:
\[ Q = (0, 0) \]
Since \(QA = 3\) is horizontal, the point \(A\) is at:
\[ A = (3, 0) \]
Since \(\angle BAQ = \frac{\pi}{2}\)., the line segment \(AB\) is vertical.
Given that \(AB = 4\) units, the point \(B\) is at:
\[ B = (3, 4) \]
Let us compute the vector \(\vec{QB}\):
\[ \vec{QB} = B - Q = (3, 4) \]
The length of \(QB\) is \(\sqrt{3^2 + 4^2} = 5\).
The slope of the line \(QB\) is \(\frac{4}{3}\).
Since \(\angle CBQ = \frac{\pi}{2}\)., the line segment \(BC\) is perpendicular to \(QB\).
The slope of \(BC\) must be the negative reciprocal of the slope of \(QB\):
\[ m_{BC} = -\frac{3}{4} \]
We are given that \(BC = 1\) unit.
Let the vector \(\vec{BC}\) be represented by \((x_{BC}, y_{BC})\).
Since the slope is \(-\frac{3}{4}\)., the unit vector along \(BC\) must point in the direction where the \(x\)-coordinate decreases and the \(y\)-coordinate increases (as seen from the geometric representation in the diagram):
\[ \vec{BC} = 1 \cdot \left(-\frac{4}{5}, \frac{3}{5}\right) = (-0.8, 0.6) \]
Now we can find the coordinates of \(C\):
\[ C = B + \vec{BC} = (3 - 0.8, 4 + 0.6) = (2.2, 4.6) \]
Since \(\angle CPQ = \frac{\pi}{2}\)., the line \(CP\) is perpendicular to the \(X\)-axis (which is the line \(QA\)).
Thus, \(P\) is the orthogonal projection of \(C\) onto the \(X\)-axis.
The \(x\)-coordinate of \(P\) must be the same as that of \(C\)., and its \(y\)-coordinate is 0:
\[ P = (2.2, 0) \]
The length of \(PQ\) is the distance from the origin \(Q(0, 0)\) to \(P(2.2, 0)\):
\[ PQ = 2.2 \text{ units} \]
• Step 4 : Final Answer:
The length of \(PQ\) is 2.2 units.
This corresponds to option (A).
