Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2\(\hat i\)-\(\hat j\)+4\(\hat k\) and is in the direction \(\hat i\)+2\(\hat j\)-\(\hat k\).
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2\(\hat i\)-\(\hat j\)+4\(\hat k\) and is in the direction \(\hat i\)+2\(\hat j\)-\(\hat k\).
Solution and Explanation
It is given that the line passes through the point with the position vector
\(\vec a\)=2\(\hat i\)-\(\hat j\)+4\(\hat k\)...(i)
\(\vec b\)=\(\hat i\)+2\(\hat j\)-\(\hat k\)....(ii)
It is known that a line through a point with position vector a→ and parallel to b→ is given by the equation,
\(\vec r\)=\(\vec a\)+λ\(\vec b\)
⇒ \(\vec r\)=2\(\hat i\)-\(\hat j\)+4\(\hat k\)+λ(\(\hat i\)+2\(\hat j\)-\(\hat k\))
This is the required equation of the line in vector form.
r→=x\(\hat i\)-y\(\hat j\)+z\(\hat k\)
⇒x\(\hat i\)-y\(\hat j\)+z\(\hat k\)=(λ+2\(\hat i\)+(2λ-1)\(\hat j\)+(-λ+4)\(\hat k\)
Eliminating, we obtain the cartesian form equation as \(\frac{x-2}{1}\)=\(\frac{y+1}{2}\)=\(\frac{z-4}{-1}\)
This is the required equation of the given line in cartesian form.
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)