Find the equation of the line that passes through the point(1,2,3)and is parallel to the vector 3\(\hat i\)+2\(\hat j\)-2\(\hat k\).
Find the equation of the line that passes through the point(1,2,3)and is parallel to the vector 3\(\hat i\)+2\(\hat j\)-2\(\hat k\).
Solution and Explanation
It is given that the line passes through the point A(1,2,3).
Therefore, the position vector through A is
\(\vec a\)=\(\hat i\)+2\(\hat j\)+3\(\hat k\), \(\vec b\)=3\(\hat i\)+2\(\hat j\)-2\(\hat k\)
It is known that the line that passes through point A and is parallel to \(\vec b\) is given by \(\vec r\)=\(\vec a\)+λ\(\vec b\), where λ is a constant.
⇒\(\hat i\)+2\(\hat j\)+3\(\hat k\)+λ+(3\(\hat i\)+2\(\hat j\)-2\(\hat k\))
This is the required equation of the line.
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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\(\hat{k}\)