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The orthogonal projection A’ of the point A with the position vector (1, 2, 3) on the plane $3x - y + 4z = 0$ isA. (-1,3,-1)B. $\left( {\dfrac{{ - 1}}{2},\dfrac{5}{2},1} \right)$C. $\left( {\dfrac{1}{2},\dfrac{{ - 5}}{2}, - 1} \right)$D. (6,-7,-5)

Last updated date: 11th Jun 2024
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Hint: We are given with a position vector and an equation of plane. So we will find the equation of line passing through the vector and perpendicular to the given plane. Then point A is on the line and it should satisfy the equations. Then it should give the required orthogonal point.

Step by step solution:
The equation of line passing through the vector and perpendicular to the plane is
$\dfrac{{x - 1}}{3} = \dfrac{{y - 2}}{{ - 1}} = \dfrac{{z - 3}}{4} = k$
These coordinates of the point lying on both and is orthogonal are $\left[ {3k + 1, - k + 2,4k + 3} \right]$
For the orthogonal projection the dot product must be zero since they are perpendicular.
$\Rightarrow 3\left( {3k + 1} \right) - 1\left( { - k + 2} \right) + 4\left( {4k + 3} \right) = 0$
On solving the equations,
$\Rightarrow 9k + 3 + k - 2 + 16k + 12 = 0$
Separating the variable terms and constants,
$\Rightarrow 26k + 13 = 0$
$\Rightarrow 26k = - 13$
To find the value of K,
$\Rightarrow k = \dfrac{{ - 1}}{2}$
Then the coordinates of the orthogonal point are given by,
$3k + 1, - k + 2,4k + 3$
Substituting the value of k,
$\Rightarrow \left[ {3\left( {\dfrac{{ - 1}}{2}} \right) + 1,\dfrac{1}{2} + 2,4\left( {\dfrac{{ - 1}}{2}} \right) + 3} \right]$
On solving the above fractions,
$\Rightarrow \left[ {\dfrac{{ - 1}}{2},\dfrac{5}{2},1} \right]$
This is the orthogonal projection of A’.

Thus option B is the correct answer.

Note:
Here note that the orthogonal is perpendicular to the given plane thus their product is taken zero. Projection is showing 3D figures in 2D form at right angles. Here the orthogonal is having the position vector on the plane given. So we first found the values of k and then that of the coordinates.