The vector equation of the plane passing through the point A ( 1 , 2 , − 1 ) and parallel to the vectors 2 i ^ + j ^ − k ^ and i ^ − j ^ + 3 k ^ is

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The vector equation of the plane passing through the point A ( 1 , 2 , − 1 ) and parallel to the vectors 2 i ^ + j ^ ​ − k ^ and i ^ − j ^ ​ + 3 k ^ is

TestHub · Accepted Answer

The normal vector to the plane is given by the cross product of the vectors parallel to the plane.

$n = (2 \hat{i} + \hat{j} - \hat{k}) \times (\hat{i} - \hat{j} + 3 \hat{k})$

$n = \hat{i} 21 \hat{j} 1 - 1 \hat{k} - 1 3$

$n = \hat{i} (3 - 1) - \hat{j} (6 + 1) + \hat{k} (- 2 - 1) = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}$

The plane passes through the point $A (1, 2, - 1)$ , so its position vector is $a = \hat{i} + 2 \hat{j} - \hat{k}$ .

The vector equation of the plane is $r \cdot n = a \cdot n$ .

$a \cdot n = (\hat{i} + 2 \hat{j} - \hat{k}) \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k})$

$a \cdot n = (1) (2) + (2) (- 7) + (- 1) (- 3) = 2 - 14 + 3 = - 9$

Therefore, the equation of the plane is $r \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = - 9$ .

Answer: $\overline{r} \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = - 9$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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