Question

Equation of the plane that contains the lines \(\mathbf{r} = (\hat{\mathbf{i}}+\hat{\mathbf{j}}) + \lambda(\hat{\mathbf{i}}+2\hat{\mathbf{j}}-\hat{\mathbf{k}})\) and \(\mathbf{r} = (\hat{\mathbf{i}}+\hat{\mathbf{j}}) + \mu(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2\hat{\mathbf{k}})\), is

• A. \(\mathbf{r} \cdot (2\hat{\mathbf{i}}+\hat{\mathbf{j}}-3\hat{\mathbf{k}}) = -4\)
• B. \(\mathbf{r} \times (-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) = 0\)
• C. \(\mathbf{r} \cdot (-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) = 0\)
• D. None of these

Hint:

The cross product of direction vectors gives the normal to the plane containing both lines.

Answer 1

The Correct Option is C

To find the equation of the plane that contains the given lines, we first need to understand the components of the problem. Let's break it down: 

1. Both given lines pass through the common point \((\hat{\mathbf{i}} + \hat{\mathbf{j}})\).
2. The direction vector of the first line is \(\mathbf{a} = (\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - \hat{\mathbf{k}})\).
3. The direction vector of the second line is \(\mathbf{b} = (-\hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}})\).

The plane containing these two lines will have a normal vector \(\mathbf{n}\) that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). Thus, the normal vector can be found using the cross product:

\(\mathbf{n} = \mathbf{a} \times \mathbf{b}\)

Computing the cross product:

\[\mathbf{n} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \\ \end{vmatrix}\]\[= \hat{\mathbf{i}} \left( (2)(-2) - (1)(-1) \right) - \hat{\mathbf{j}} \left( (1)(-2) - (-1)(-1) \right) + \hat{\mathbf{k}} \left( (1)(1) - (-1)(2) \right)\]\[= \hat{\mathbf{i}}(-4 + 1) - \hat{\mathbf{j}}(-2 - 1) + \hat{\mathbf{k}}(1 + 2)\]\[= -3\hat{\mathbf{i}} + 3\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\]

Thus, the normal vector \(\mathbf{n} = -3\hat{\mathbf{i}} + 3\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\), which can be simplified to \(-\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}\).

Now, using the point \((\hat{\mathbf{i}} + \hat{\mathbf{j}})\) which lies on the plane, the equation of the plane is:

\[\begin{align*} & \mathbf{n} \cdot \left(\mathbf{r} - (\hat{\mathbf{i}} + \hat{\mathbf{j}})\right) = 0 \\ & \Rightarrow \left(-\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}\right) \cdot \left(x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}} - \hat{\mathbf{i}} - \hat{\mathbf{j}}\right) = 0 \\ & \Rightarrow -x + y + z - (-1) + 1 + 0 = 0 \\ & \Rightarrow -x + y + z = 0 \end{align*}\]

This can also be expressed as \(\mathbf{r} \cdot (-\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) = 0\).

We can now conclude that the correct answer is:

\(\mathbf{r} \cdot (-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) = 0\)