Direction Cosines & Lines

1 Mark Questions

1. Write the distance of a point <math>P(a,b,c)</math> from
2. axis. Delhi 2014C

Let any point on X-axis be <math>Q(x, 0, 0)</math>. Then, use the formula for distance of point <math>R(x_1, y_1, z_1)</math> from <math>S(x_2, y_2, z_2)</math> <math display="block">= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}</math>

Given point is <math>P(a, b, c)</math>.

Let the coordinates of the point on X-axis be <math>(a,0,0)</math>. (1/2)

[: x-coordinate of both points will be same]

.: Required distance

<math display="block">= \sqrt{(a-a)^2 + (0-b)^2 + (0-c)^2}</math>

<math>=\sqrt{0+b^2+c^2}</math>

<math display="block">=\sqrt{b^2+c^2}</math>

(1/2)

1. If the cartesian equation of a line is <math>\frac{3-x}{5} = \frac{y+4}{7} = \frac{2\dot{z}-6}{4}</math>, then write the vector

equation for the line. All India 2014

Given cartesian equation of a line is

<math display="block">\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}</math>

On rewriting the given equation in standard form, we get

[let]

<math display="block">\frac{x-3}{-5} = \frac{y+4}{7} = \frac{z-3}{2} = \lambda</math>

<math>x = -5\lambda + 3</math>, <math>y = 7\lambda - 4</math>

<math>\Rightarrow</math> and <math>z = 2\lambda + 3</math>

(1/2)

Now, <math>x\hat{i} + y\hat{j} + z\hat{k}</math>

<math>=(-5\lambda + 3)\hat{i} + (7\lambda - 4)\hat{j} + (2\lambda + 3)\hat{k}</math>

<math display="block">=3\hat{i}-4\hat{j}+3\hat{k}+\lambda(-5\hat{i}+7\hat{j}+3\hat{k})</math>

<math display="block">\Rightarrow \vec{r} = (3\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(-5\hat{i} + 7\hat{j} + 3\hat{k})</math>

which is the required equation of line in vector form. (1/2)

1. Write the equation of the straight line through the point <math>(\alpha, \beta, \gamma)</math> and parallel to
2. axis.