CUET Preparation Today
Find the vector and the cartesian equations of a line that passes through the point $A(1, 2, -1)$ and parallel to the line $5x - 25 = 14 - 7y = 35z$. |
$\vec{r} = (5\hat{i} + 2\hat{j}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k})$; $\frac{x-5}{7} = \frac{y-2}{-5} = \frac{z}{1}$ $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k})$; $\frac{x-1}{7} = \frac{y-2}{-5} = \frac{z+1}{1}$ $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(5\hat{i} - 7\hat{j} + 35\hat{k})$; $\frac{x-1}{5} = \frac{y-2}{-7} = \frac{z+1}{35}$ $\vec{r} = (7\hat{i} - 5\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$; $\frac{x-7}{1} = \frac{y+5}{2} = \frac{z-1}{-1}$ |
$\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k})$; $\frac{x-1}{7} = \frac{y-2}{-5} = \frac{z+1}{1}$ |
The correct answer is Option (2) → $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k})$; $\frac{x-1}{7} = \frac{y-2}{-5} = \frac{z+1}{1}$ ## Given point $A(1, 2, -1)$ Given line $5x - 25 = 14 - 7y = 35z$, $\Rightarrow 5(x - 5) = -7(y - 2) = 35z$, $\Rightarrow \frac{x-5}{7} = \frac{y-7}{-5} = \frac{z}{1}$ [Divide by 35] Direction ratio of the line $(7, -5, 1)$ $∴$ Direction ratio of the parallel line $(7, -5, 1)$ Equation of the line passing through the point $A(1, 2, -1)$ and parallel to the given line $\frac{x-1}{7} = \frac{y-2}{-5} = \frac{z+1}{1}$ Vector form of the line $\hat{i} + 2\hat{j} - \hat{k} + \lambda(7\hat{i} - 5\hat{j} + \hat{k})$ |
