Question

Line \(L_1\) passes through the points \(\mathbf{i} + \mathbf{j}\) and \(\mathbf{k} - \mathbf{i}\). Line \(L_2\) passes through the point \(\mathbf{j} + 2\mathbf{k}\) and is parallel to the vector \(\mathbf{i} + \mathbf{j} + \mathbf{k}\). If \(\mathbf{x}i + \mathbf{y}j + \mathbf{z}k\) is the point of intersection of the lines \(L_1\) and \(L_2\), then find \((y - x) =\)

• A. \(2z\)
• B. \(-2z\)
• C. \(z\)
• D. \(-z\)

Hint:

Use parametric equations of lines and solve for intersection points by equating coordinates. Substitute back to find required expressions.

Answer 1

The Correct Option is C

Parametric form of \(L_1\): Let \(L_1\) pass through \(\mathbf{i} + \mathbf{j}\) with direction vector \(\mathbf{k} - \mathbf{i} - (\mathbf{i} + \mathbf{j}) = -2\mathbf{i} - \mathbf{j} + \mathbf{k}\). So, \[ \mathbf{r} = (\mathbf{i} + \mathbf{j}) + t(-2\mathbf{i} - \mathbf{j} + \mathbf{k}). \] Parametric form of \(L_2\): \[ \mathbf{r} = (\mathbf{j} + 2\mathbf{k}) + s(\mathbf{i} + \mathbf{j} + \mathbf{k}). \] Equate coordinates: \[ x: 1 - 2t = s,
y: 1 - t = 1 + s,
z: t = 2 + s. \] From the second equation: \[ 1 - t = 1 + s \implies -t = s \implies s = -t. \] From the first equation: \[ 1 - 2t = s = -t \implies 1 - 2t = -t \implies 1 = t. \] From the third: \[ z = t = 1. \] Calculate \(y - x\): \[ y = 1 - t = 1 - 1 = 0,
x = 1 - 2t = 1 - 2 = -1, \] so \[ y - x = 0 - (-1) = 1 = z. \] Hence, \(y - x = z\).