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A Line Passes Through the Point with Position Vector 2 ^ I − 3 ^ J + 4 ^ K and is in the Direction of 3 ^ I + 4 ^ J − 5 ^ K . Find Equations of the Line in Vector and Cartesian Form. - Mathematics

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प्रश्न

A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of  \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form. 

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उत्तर

We know that the vector equation of a line passing through a point with position vector `vec a` and parallel to the vector `vec b` is \[\vec{r} = \vec{a} + \lambda \vec{b}\]

Here,

\[\vec{a} = 2 \hat {i} - 3 \hat{j} + 4 \hat {k} \]

\[ \vec{b} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}\]

So, the vector equation of the required line is

\[\vec{r} = \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda \left( 3 \hat{i} + 4 \hat{j} - 5 \hat{k} \right) . . . (1)\]

\[\text{ Here }, \lambda \text { is a parameter }. \]

Reducing (1) to cartesian form, we get

\[x \hat{i} + y \hat{j} + z \hat{k} = \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda \left( 3 \hat{i} + 4 \hat{j} - 5 \hat{k} \right) [\text{ Putting } \vec{r} = x \hat{i} + y \hat{j}+ z \hat{k} \text{ in } (1)]\]

\[ \Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = \left( 2 + 3\lambda \right) \hat{i} + \left( - 3 + 4\lambda \right) \hat{j} + \left( 4 - 5\lambda \right) \hat{k} \]

\[\text{ Comparing the coefficients of }  \hat{i} , \hat{j} \text{ and  } \hat{k} , \text{ we get } \]

\[x = 2 + 3\lambda, y = - 3 + 4\lambda, z = 4 - 5\lambda\]

\[ \Rightarrow \frac{x - 2}{3} = \lambda, \frac{y + 3}{4} = \lambda, \frac{z - 4}{- 5} = \lambda\]

\[ \Rightarrow \frac{x - 2}{3} = \frac{y + 3}{4} = \frac{z - 4}{- 5} = \lambda\]

\[\text{ Hence, the cartesian form of (1) is } \]

\[\frac{x - 2}{3} = \frac{y + 3}{4} = \frac{z - 4}{- 5}\]

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अध्याय 28: Straight Line in Space - Exercise 28.1 [पृष्ठ ९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.1 | Q 4 | पृष्ठ ९

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