The projection of point Pp- on the plane r-n-q is s- then-

The correct option is B s⃗=p⃗+(q p⃗·n⃗)n⃗|n⃗|^2 The equation of line passing through p⃗ and perpendicular to plane be r⃗.n⃗=q is nbsp; r ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Parametric Equation of Parabola

The projectio...

Question

The projection of point $P (\to p)$ on the plane $\to r \cdot \to n = q$ is $(\to s)$ , then:

\to s = \frac{(q - \to p \cdot \to n) \to n}{∣ \to n ∣^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\to s = \to p + \frac{(q - \to p \cdot \to n) \to n}{∣ \to n ∣^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\to s = \to p - \frac{(\to p \cdot \to n) \to n}{∣ \to n ∣^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\to s = \to p - \frac{(q - \to p \cdot \to n) \to n}{∣ \to n ∣^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\to p

\to s

\to r \times \to p = \to q \times \to p

\to r . \to s = 0,

\to r =

\to P + \to Q = \to R a n d \to P - \to Q = \to S,

R^{2} + S^{2} = (P^{2} + Q^{2})

\to p = 3 \to a - 5 \to b; \to q = 2 \to a + \to b; \to r = \to a + 4 \to b; \to s = - \to a + \to b

sin (\to p \land \to q) = 1

sin (\to r \land \to s) = 1

cos (\to a \land \to b)

\to p = 5 \to a - 3 \to b; \to q = - \to a - 2 \to b

\to r = - 4 \to a - \to b; \to s = - \to a + \to b

\to x = \frac{1}{3} (\to p + \to r + \to s)

\to y = \frac{1}{5} (\to r + \to s)

\to p, \to q

\to r

R^{3}

\to s

\to p, \to q

\to r

4

3

5

\to s

(- \to p + \to q + \to r)

(\to p - \to q + \to r)

(- \to p - \to q + \to r)

x, y

z

2 x + y + z

Join BYJU'S Learning Program