Question

The direction cosine of \(\mathrm{x}=\mathrm{ay}+\mathrm{b}, \mathrm{z}=\mathrm{cy}+\mathrm{d}\) (A) \(\pm\left[a / \sqrt{\left(a^{2}+c^{2}+1\right)}\right], \pm\left[1 / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left.\pm\left[\mathrm{c} / \sqrt{(} \mathrm{a}^{2}+\mathrm{c}^{2}+1\right)\right]\) (C) \(\left[(-a) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right],\left[(-1) / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left[(-c) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right]\) (D) None of these

Short answer

(A) \(\pm\left[a / \sqrt{\left(a^{2}+c^{2}+1\right)}\right], \pm\left[1 / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left.\pm\left[\mathrm{c} / \sqrt{(} \mathrm{a}^{2}+\mathrm{c}^{2}+1\right)\right]\)

Step-by-step solution

Write the Equations in Parametric Form

The given equations are \(x=a y+b\) and \(z=c y+d\). Let's write these equations in parametric form. First, let's choose \(y\) as the parameter \(t\). Then, we have $$x=a\ t+b$$ $$y=t$$ $$z=c\ t+d$$

Calculate Direction Vector

Now, we have the parametric equations. From these equations, we can find the direction vector, which is the derivative of the position vector with respect to the parameter \(t\). Let's denote the position vector \(\overrightarrow{r}=x\ \hat{i}+y\ \hat{j}+z\ \hat{k}\), and its derivative with respect to \(t\) as \(\overrightarrow{r'}\). $$\overrightarrow{r'}=\frac{d \overrightarrow{r}}{d t}=\left(\frac{d x}{d t}\right) \hat{i}+\left(\frac{d y}{d t}\right) \hat{j}+\left(\frac{d z}{d t}\right) \hat{k}$$ Taking the derivatives with respect to \(t\) yields: $$\overrightarrow{r'}=a\ \hat{i}+\hat{j}+c\ \hat{k}$$

Calculate the Magnitude of the Direction Vector

To normalize the direction vector, we need to calculate its magnitude. The magnitude of the direction vector \(\overrightarrow{r'}\) is given by: $$|\overrightarrow{r'}|=\sqrt{(a)^2+(1)^2+(c)^2}$$ $$|\overrightarrow{r'}|=\sqrt{a^2+1+c^2}$$

Normalize the Direction Vector to Get Direction Cosines

Next, we normalize the direction vector by dividing each component of the vector with the magnitude of the vector. The direction cosine vector \(\overrightarrow{l}\) is equal to: $$\overrightarrow{l}=\frac{\overrightarrow{r'}}{|\overrightarrow{r'}|}$$ $$\overrightarrow{l}=\left[\frac{a}{\sqrt{a^2+1+c^2}}\right] \hat{i}+\left[\frac{1}{\sqrt{a^2+1+c^2}}\right] \hat{j}+\left[\frac{c}{\sqrt{a^2+1+c^2}}\right] \hat{k}$$ Comparing the components with answer choices, we find that our solution is equal to option (A): (A) \(\pm\left[a / \sqrt{\left(a^{2}+c^{2}+1\right)}\right], \pm\left[1 / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left.\pm\left[\mathrm{c} / \sqrt{(} \mathrm{a}^{2}+\mathrm{c}^{2}+1\right)\right]\)

Key concepts

Parametric Equations

Parametric equations allow us to express the components of mathematical objects in terms of an independent parameter. In this context, they help express a line in three-dimensional space. The given equations \( x = ay + b \) and \( z = cy + d \) can both be rewritten using a parameter \( t \). By choosing \( y = t \), we convert the points into parametric equations:

• \( x = at + b \)
• \( y = t \)
• \( z = ct + d \)

These equations describe each coordinate along a path as a function of \( t \). This method greatly simplifies the processes of finding direction vectors and performing vector operations.

Direction Vector

The direction vector of a line, particularly in 3D space, indicates the direction along which the line extends. It's obtained by differentiating the components of the position vector \( \overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k} \) with respect to the parameter \( t \). For the parametric equations \( x = at + b \), \( y = t \), and \( z = ct + d \), taking the derivative of each with respect to \( t \) provides:

• \( \frac{dx}{dt} = a \)
• \( \frac{dy}{dt} = 1 \)
• \( \frac{dz}{dt} = c \)

Thus, the direction vector \( \overrightarrow{r'} \) becomes:\[ \overrightarrow{r'} = a \hat{i} + \hat{j} + c \hat{k} \] This vector highlights how changes in \( t \) influence the coordinates of points along the line.

Vector Normalization

Normalization takes a non-zero vector and scales it to have unit length, making it a direction vector known as direction cosines. The first step requires computing the vector's magnitude (which is discussed in the next section). For the direction vector \( \overrightarrow{r'} = a \hat{i} + \hat{j} + c \hat{k} \), normalize it by dividing each component by its magnitude \( |\overrightarrow{r'}| \). Thus, the normalized form is given by:\[\overrightarrow{l} = \frac{a}{\sqrt{a^2 + 1 + c^2}} \hat{i} + \frac{1}{\sqrt{a^2 + 1 + c^2}} \hat{j} + \frac{c}{\sqrt{a^2 + 1 + c^2}} \hat{k}\]Direction cosines reflect the proportional direction of vectors relative to the axes and are vital in geometry, physics, and engineering.

Magnitude of a Vector

The magnitude, or length, of a vector gives us its size regardless of direction. It’s calculated using the square root of the sum of the squares of its components. For a vector \( \overrightarrow{v} = a \hat{i} + b \hat{j} + c \hat{k} \), the magnitude is:\[|\overrightarrow{v}| = \sqrt{a^2 + b^2 + c^2}\]In our case, the direction vector \( \overrightarrow{r'} = a \hat{i} + \hat{j} + c \hat{k} \) has a magnitude:\[|\overrightarrow{r'}| = \sqrt{a^2 + 1^2 + c^2} = \sqrt{a^2 + 1 + c^2}\]Calculating this magnitude is crucial for normalizing the vector, ensuring all vector components retain direction but scale down to fit within a unit length. This helps in standardizing vectors to compare their directions without the influence of their scale.