Question
Download Solution PDFThe equation of the plane that passes through (1, -1, 2) and has direction ratios (1, 2, 3) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The Plane passes through the point having a position vector \(\overrightarrow{a}=i-j+2k\) and is the normal vector \(\overrightarrow{n}=i+2j+3k\)
So, the vector equation of the plane is
\((\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\)
⇒ \(\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\)
⇒ \(\overrightarrow{r}.(i+2j+3k)=(i-j+2k).(i+2j+3k)\)
Let \(\overrightarrow{r}=xi+yj+zk\)
⇒ \((xi+yj+zk).(i+2j+3k)=(i-j+2k).(i+2j+3k)\)
⇒ x + 2y + 3z = 1 - 2 + 6
⇒ x + 2y + 3z = 5
Hence, the cartesian equation of the plane is x + 2y + 3z = 5.
Last updated on May 26, 2025
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