Question

If A has coordinates \((-4,2,1)\) and \(\mathrm{B}\) has coordinates \((2,0,2)\) find the direction ratio of the vector \(\overrightarrow{\mathrm{AB}}\). Find its direction cosines \(l, m\) and \(n\) and verify that \(l^{2}+m^{2}+n^{2}=1\).

Short answer

Question: Given points A(-4, 2, 1) and B(2, 0, 2), find the direction cosines of the vector AB and verify that \(l^2 + m^2 + n^2 = 1\). Solution: The direction cosines of the vector AB are: \(l = \frac{6}{\sqrt{41}}\), \(m = \frac{-2}{\sqrt{41}}\), \(n = \frac{1}{\sqrt{41}}\). Verifying that \(l^2 + m^2 + n^2 = 1\), we find that it holds true.

Step-by-step solution

Finding the direction ratio of the vector \(\overrightarrow{AB}\)

Subtract the coordinates of point A from point B: \(x = 2 - (-4) = 6\) \(y = 0 - 2 = -2\) \(z = 2 - 1 = 1\) So, the direction ratio of the vector \(\overrightarrow{AB}\) is \((6, -2, 1)\).

Finding the magnitude of the vector \(\overrightarrow{AB}\)

We can find the magnitude of the vector by using the formula: \(magnitude = \sqrt{x^2 + y^2 + z^2}\) \(magnitude = \sqrt{6^2 + (-2)^2 + 1^2} = \sqrt{36 + 4 + 1} = \sqrt{41}\)

Finding the direction cosines \(l, m\), and \(n\)

We can find the direction cosines using the formula: \(l = \frac{x}{\sqrt{x^2 + y^2 + z^2}}\), \(m = \frac{y}{\sqrt{x^2 + y^2 + z^2}}\), \(n = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\) \(l = \frac{6}{\sqrt{41}}\), \(m = \frac{-2}{\sqrt{41}}\), \(n = \frac{1}{\sqrt{41}}\) So, the direction cosines are: \(l = \frac{6}{\sqrt{41}}\), \(m = \frac{-2}{\sqrt{41}}\), \(n = \frac{1}{\sqrt{41}}\)

Verifying that \(l^2 + m^2 + n^2 = 1\)

We will now square each direction cosine and sum them up: \(l^2 + m^2 + n^2 = \left(\frac{6}{\sqrt{41}}\right)^2 + \left(\frac{-2}{\sqrt{41}}\right)^2 + \left(\frac{1}{\sqrt{41}}\right)^2\) \(l^2 + m^2 + n^2 = \frac{36}{41} + \frac{4}{41} + \frac{1}{41}\) \(l^2 + m^2+ n^2 = \frac{36 + 4 + 1}{41}\) \(l^2 + m^2 + n^2 = \frac{41}{41}\) \(l^2 + m^2 + n^2 = 1\) Thus, we have verified that \(l^2 + m^2 + n^2 = 1\).

Key concepts

Vector Algebra

Vector algebra is an essential area of mathematics that deals with entities known as vectors. Vectors have both magnitude and direction, distinguishing them from simple numbers which only have magnitude. When we talk about vectors in a three-dimensional space, we are usually referring to '3D vectors' that have three components.

In our exercise, the vector \(\overrightarrow{AB}\) represents the direction and distance from point A to point B in a 3D space. The 'direction ratios' give us a proportional relationship between the x, y, and z components of this vector. Finding the direction ratios is the first step in understanding the behavior of \(\overrightarrow{AB}\), and it's achieved through subtracting the coordinates of the initial point from the final point. This action gives us a new vector that points from A to B, providing valuable geometrical insights.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, combines algebra and geometry to describe the position of points on a plane using ordered pairs (like (x, y)) or triples (like (x, y, z)) for two and three dimensions respectively. It is the backdrop against which we calculate direction ratios and cosines in vector analysis.

In the context of our problem, point A located at \( (-4,2,1) \) and point B at \( (2,0,2) \) lie within a three-dimensional coordinate system. Using the Cartesian plane, we decipher the underlying spatial structure and manipulate points and vectors algebraically. By subtracting the coordinates of point A from B, we find the components of the vector \(\overrightarrow{AB}\), which leads directly into the concept of direction cosines.

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size, which is a scalar quantity. For a vector in three-dimensional space, the magnitude is found by taking the square root of the sum of the squares of its components, depicted as \( magnitude = \sqrt{x^2 + y^2 + z^2} \).

When we calculate the magnitude of vector \(\overrightarrow{AB}\), we're seeking the straight-line distance between points A and B in the three-dimensional space. This step is critical before we can determine the direction cosines because they require the components of the vector to be normalized by the vector's magnitude, symbolizing how much the vector 'leans' into each axis of the three-dimensional plane. This resulting set of direction cosines is a normalized form of the direction ratios and will always satisfy the condition \( l^2 + m^2 + n^2 = 1 \) which is fundamental to vector algebra.