Question

Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.

Answer 1

The given equation of the plane is

\[12x - 3y + 4z = 1\]

` ⇒  \left( x\hat{ i }  +y \hat{ j }+ z \hat{ k}  \right) . \left( \text{ 12}  \hat{ i }  - \text{ 3 }\hat{ j }  + 4 \hat{ k }  \right) = 1 `

`    ⇒  \vec{r} . \left( \text{ 12 } \hat{ i }  - \text{ 3 }\hat{ j } + 4\hat{ k} \right) = 1,\text{  which is the vector equation of the plane.}` 

`\text{ Because the vector equation of the plane is } \vec{r} . \vec{n} = \vec{a} . \vec{n} )`

`\text{ So, the normal vector, }\vec{n} = \text{ 12 }\hat{ i } - \text{ 3 }\hat{ j } + 4\hat{ k }`

\[\left| \vec{n} \right| = \sqrt{144 + 9 + 16} = 13\]

`\text{ Unit vector parallel to }\vec{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{\text{ 12 }\hat{ i } - \text{ 3 }\hat{ j } + 4 \hat{ k }}{13}`

So, the vector of magnitude 26 units normal to the plane

`= 26 \times \frac{\text{ 12 } \hat{ i } - \text{ 3 }\hat{ j} + 4\hat{ k }}{13}`

`= 2 \left( \text{ 12 }\hat{ i }- \text{ 3 }\hat{ j } + 4 \hat{ k } )`

`= \text{ 24 }\hat{ i } - \text{ 6 }\hat{ i } + 8 \hat{ k } `

\[\]

\[\]