If the position vector \({\rm{\vec a}}\) of the point (5, n) is such that \(\left| {{\rm{\vec a}}} \right| = 13\) , then the value/values of n can be
± 12
Understanding Position Vectors and Magnitude
A position vector represents the location of a point in space relative to the origin (0,0). For a point (x, y), the position vector can be written as \(\vec r = x\hat i + y\hat j\), where \(\hat i\) and \(\hat j\) are unit vectors along the x and y axes, respectively.
The magnitude of a position vector \(\vec r = x\hat i + y\hat j\) (or point (x, y)) is the distance from the origin to the point. It is calculated using the Pythagorean theorem:
\[\left| {\vec r} \right| = \sqrt{x^2 + y^2}\]
Solving the Position Vector Problem
We are given that the position vector \({\rm{\vec a}}\) is for the point (5, n). This means the components of the vector are 5 and n. We can write the position vector as \({\rm{\vec a}} = 5\hat i + n\hat j\).
We are also given that the magnitude of this vector is \(\left| {{\rm{\vec a}}} \right| = 13\).
Using the formula for the magnitude of a vector, we have:
\[\left| {{\rm{\vec a}}} \right| = \sqrt{5^2 + n^2}\]
We are given \(\left| {{\rm{\vec a}}} \right| = 13\), so we can set up the equation:
\[\sqrt{5^2 + n^2} = 13\]
Finding the Value(s) of n
To solve for n, we need to eliminate the square root. We do this by squaring both sides of the equation:
\[\left(\sqrt{5^2 + n^2}\right)^2 = 13^2\]
\[5^2 + n^2 = 169\]
Now, we calculate the value of \(5^2\):
\[25 + n^2 = 169\]
Next, isolate \(n^2\) by subtracting 25 from both sides:
\[n^2 = 169 - 25\]
\[n^2 = 144\]
Finally, to find n, we take the square root of both sides. Remember that taking the square root of a positive number yields both a positive and a negative result:
\[n = \pm \sqrt{144}\]
\[n = \pm 12\]
Thus, the possible values for n are 12 and -12.
Conclusion on the Value(s) of n
Based on our calculation, the value/values of n can be \(\pm 12\).
Revision Table: Position Vector and Magnitude
| Concept | Description | Formula (for 2D vector \((x, y)\)) |
|---|---|---|
| Position Vector | Vector from the origin (0,0) to a point (x, y). | \(\vec r = x\hat i + y\hat j\) or \((x, y)\) |
| Magnitude of a Vector | The length or size of the vector. Distance from origin for a position vector. | \(\left| {\vec r} \right| = \sqrt{x^2 + y^2}\) |
Additional Information on Vector Components
In the given problem, the point (5, n) corresponds to the position vector \({\rm{\vec a}}\). The 'x' component of the vector is 5, and the 'y' component is n. The magnitude calculation \(\sqrt{5^2 + n^2}\) comes directly from the Pythagorean theorem applied to a right triangle formed by the origin, the point (5, 0), and the point (5, n). The sides of this triangle have lengths 5 (along the x-axis), |n| (along the y-axis), and the hypotenuse is the magnitude of the vector, which is 13.
The fact that \(n^2 = 144\) means that n can be either 12 or -12 because squaring either 12 or -12 results in 144. Both points (5, 12) and (5, -12) are 13 units away from the origin (0,0).
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