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Question Number 73782 by liki last updated on 15/Nov/19

Commented by liki last updated on 15/Nov/19

$I n e e d h e l p p l z t h a t q u e s t i o n v e c t o r . . . .$
Commented by liki last updated on 15/Nov/19

$. . . p l z M r m i n d i s p o w e r i n e e d h e l p t h a t q n o f v e c t o r$ $(a), (b) a n d (c) o r a n y o n e t o a s s i s t m e p l z$
	Answered by mind is power last updated on 15/Nov/19
	$a) let A(1,1,1),B(2,3,−1) is S∣ interesection of (AB) withe plan (p) x+y+2z−5=0 AB^→ (1,2,−2) parametric equation x(t)=t+1,y(t)=2t+1,z(t)=−2t+1 S∈(p)∩(AB)⇔(t+1)+(2t+1)+2(−2t+1)=5 ⇔−t+4=5⇒t=−1⇒S(0,−1,3) b) let (d) bee this line we haven^→ (5,1,2)⊥(p)(5x+y+2z=7) ⇒vector director of d // n^→ (x,y,z)∈d⇔∃t∈R∣(x(t)=5t+1,y(t)=t+2,z(t)=2t+1} parametric equation t=y−2 ⇒ { ((x−5y+9=0)),((z−2y+3=0)) :} c)Volum=∣a^→ .(b^→ ∧c^→ )∣ b∧c=−28i^→ +30j^→ +6k^→ ∣a^→ .(b∧c)∣=∣2.−28+4.30+6.1∣=70 ua^3$
	$a) let A (1, 1, 1), B (2, 3, - 1)$ $is S ∣ interesection of (AB) withe plan (p) x + y + 2 z - 5 = 0$ $A \vec{B} (1, 2, - 2)$ $parametric equation$ $x (t) = t + 1, y (t) = 2 t + 1, z (t) = - 2 t + 1$ $S \in (p) \cap (AB) \Leftrightarrow (t + 1) + (2 t + 1) + 2 (- 2 t + 1) = 5$ $\Leftrightarrow - t + 4 = 5 \Rightarrow t = - 1 \Rightarrow S (0, - 1, 3)$ $b) let (d) bee this line we have \vec{n} (5, 1, 2) ⊥ (p) (5 x + y + 2 z = 7)$ $\Rightarrow vector director of d / / \vec{n}$ $(x, y, z) \in d \Leftrightarrow \exists t \in R ∣ (x (t) = 5 t + 1, y (t) = t + 2, z (t) = 2 t + 1}$ $parametric equation$ $t = y - 2 \Rightarrow {\begin{cases} x - 5 y + 9 = 0 \\ z - 2 y + 3 = 0 \end{cases}$ $c) Volum =∣ \vec{a} . (\vec{b} \land \vec{c}) ∣$ $b \land c = - 28 \vec{i} + 30 \vec{j} + 6 \vec{k}$ $∣ \vec{a} . (b \land c) ∣=∣ 2 . - 28 + 4.30 + 6.1 ∣= 70 {ua}^{3}$
	Commented by liki last updated on 15/Nov/19

	$. . . T h a n k s v e r y m u c h s i r i a p p r i c i a t e y o u r w o r k!$
	Commented by mind is power last updated on 15/Nov/19

	$y^{'} re welcom$
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