The projection of point P ( vec p ) on the plane vec r ⋅ vec n = q is ( vec s ), then

Question

Tardigrade Admin · Accepted Answer

We have vec s - vec p =λ vec n and vec s ⋅ vec n = q. Thus, (λ vecn+ vecp) ⋅ vecn= q or λ=( q - vec p ⋅ vec n /| vec n |2) ⇒ vec s = vec p +(( q - vec p ⋅ vec n ) vec n /| vec n |2)

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

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

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

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

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

We have $s - p = λ n$ and $s \cdot n = q$ .
Thus, $(λ n + p) \cdot n = q$
or $λ = \frac{q - p \cdot n}{∣ n ∣ ^{2}}$
$\Rightarrow s = p + \frac{( q - p \cdot n ) n}{∣ n ∣ ^{2}}$

Q. The projection of point P(p​) on the plane r⋅n=q is (s), then

s=∣n∣2(q−p​⋅n)n​

s=p​+∣n∣2(q−p​⋅n)n​

s=p​−∣n∣2(p​⋅n)⋅n​

s=p​−∣n∣2(q​−p​⋅n)⋅n​

We have s−p​=λn and s⋅n=q. Thus, (λn+p​)⋅n=q or λ=∣n∣2q−p​⋅n​ ⇒s=p​+∣n∣2(q−p​⋅n)n​

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

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

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

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

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

We have $s - p = λ n$ and $s \cdot n = q$ .
Thus, $(λ n + p) \cdot n = q$
or $λ = \frac{q - p \cdot n}{∣ n ∣ ^{2}}$
$\Rightarrow s = p + \frac{( q - p \cdot n ) n}{∣ n ∣ ^{2}}$