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Home » A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.
A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.
CBSE | Class 12 | Exercise 27A | Maths | RS Aggarwal | Straight Line in Space
A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.

Answer
Given: line passes through point (3,4,5) and is parallel to 2 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}
To find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}
Cartesian form: \frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~b}_{3}}=\lambda
where \overrightarrow{\mathrm{a}}=\mathrm{x}_{1} \hat{\imath}+\mathrm{y}_{1} \hat{\jmath}+\mathrm{z}_{1} \hat{\mathrm{k}} is a point on the line and \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}} is a vector parallel to the line.
Explanation:
Here, \overrightarrow{\mathrm{a}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{\mathrm{k}} and \overrightarrow{\mathrm{b}}=2 \hat{\imath}+2 \hat{\jmath}-3 \hat{\mathrm{k}}
Therefore,
VeAnswer
Given: line passes through point (3,4,5) and is parallel to 2 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}
To find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}
Cartesian form: \frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~b}_{3}}=\lambda
where \overrightarrow{\mathrm{a}}=\mathrm{x}_{1} \hat{\imath}+\mathrm{y}_{1} \hat{\jmath}+\mathrm{z}_{1} \hat{\mathrm{k}} is a point on the line and \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}} is a vector parallel to the line.
Explanation:
Here, \overrightarrow{\mathrm{a}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{\mathrm{k}} and \overrightarrow{\mathrm{b}}=2 \hat{\imath}+2 \hat{\jmath}-3 \hat{\mathrm{k}}

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