Vector Calculus – Page 14

Show-that-the-plane-2x-2y-z-12-0-touches-the-sphere-x-2-y-2-z-2-2x-4y-2z-3-0-Find-the-point-of-contact-

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Question Number 133445 by benjo_mathlover last updated on 22/Feb/21 $Show that the plane 2 x - 2 y + z + 12 = 0$ $touches the sphere x^{2} + y^{2} + z^{2} - 2 x - 4 y + 2 z - 3 = 0$ $Find the point of contact .$ Answered by MJS_new last updated on…

Question-133392

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Question Number 133392 by rexford last updated on 21/Feb/21 Commented by EDWIN88 last updated on 22/Feb/21 $qn 133398$ Terms of Service Privacy…

Question-133398

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Question Number 133398 by rexford last updated on 21/Feb/21 Answered by EDWIN88 last updated on 22/Feb/21 $$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}+\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}=\left(\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{c}}\right).\overset{\rightarrow} {{b}}=−\overset{\rightarrow} {{b}}.\overset{\rightarrow}…

Question-133282

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Question Number 133282 by rexford last updated on 20/Feb/21 Answered by mr W last updated on 25/Feb/21 $(b \times c) \times a$ $= [(1, 2, - 1) \times (1, 1, - 2)] \times (2, - 1, 1)$ $= (- 3, 1, - 1) \times (2, - 1, 1)$ $$=\left(\mathrm{0},\mathrm{1},\mathrm{1}\right) \…

Question-133240

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Question Number 133240 by rexford last updated on 20/Feb/21 Answered by Kunal12588 last updated on 20/Feb/21 $r \cos γ = 2$ $\cos γ = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$ $γ = \cos^{- 1} \frac{2}{3}$ Answered…

Question-133206

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Question Number 133206 by rexford last updated on 20/Feb/21 Answered by EDWIN88 last updated on 20/Feb/21 $i + j + 3 k = (\begin{matrix} 1 \\ 1 \\ 3 \end{matrix}); 3 i - 3 j + k = (\begin{matrix} 3 \\ - 3 \\ 1 \end{matrix})$ $- 4 i + 5 j = (\begin{matrix} - 4 \\ 5 \\ 0 \end{matrix})$ $\Leftrightarrow (\begin{matrix} 1 \\ 1 \\ 3 \end{matrix}) x + (\begin{matrix} 3 \\ - 3 \\ 1 \end{matrix}) y + (\begin{matrix} - 4 \\ 5 \\ 0 \end{matrix}) z = λ (\begin{matrix} x \\ y \\ z \end{matrix})$ $\Leftrightarrow (\begin{matrix} x + 3 y - 4 z \\ x - 3 y + 5 z \\ 3 x + y + 0 z \end{matrix}) = λ (\begin{matrix} x \\ y \\ z \end{matrix})$ $$\Leftrightarrow\: $(\begin{matrix} 1 3 - 4 \\ 1 - 3 5 \\ 3 1 0 \end{matrix})$ \: $(\begin{matrix} x \\ y \\ z \end{matrix})$ \:=\:\lambda\: $(\begin{matrix} x \\ y \\ z \end{matrix})$ …

Is-the-following-series-absolutely-convergent-S-1-n-1-1-n-n-1-Is-the-following-series-absolutely-convergent-S-2-n-1-1-n-1-n-1-

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Question Number 2045 by prakash jain last updated on 31/Oct/15 $I s t h e f o l l o w i n g s e r i e s a b s o l u t e l y c o n v e r g e n t ?$ $S_{1} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 1)}$ $I s t h e f o l l o w i n g s e r i e s a b s o l u t e l y c o n v e r g e n t ?$ $S_{2} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1})$ …

proof-that-2-is-an-irrational-number-

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Question Number 1865 by Denbang last updated on 18/Oct/15 $p r o o f t h a t \sqrt{2} i s a n i r r a t i o n a l n u m b e r$ Answered by 123456 last updated on 18/Oct/15 $suppuse by absurf that \sqrt{2} \in Q, then$ $$\exists\left({p},{q}\right)\in\mathbb{Z},{q}\neq\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}},\left({p},{q}\right)=\mathrm{1} \…

Find-the-angle-between-the-line-x-2-3-y-1-1-z-3-2-and-the-plane-3x-4y-z-5-0-

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Question Number 132904 by bramlexs22 last updated on 17/Feb/21 $Find the angle between the line$ $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2} and the$ $plane 3 x + 4 y + z + 5 = 0$ Answered by mr W last updated on 17/Feb/21 $${l}:\:\left(\mathrm{3},−\mathrm{1},\:\mathrm{2}\right)…

Show-by-use-of-the-characteristics-of-a-vector-space-that-the-set-x-R-with-the-following-operations-and-builds-a-vector-space-Use-vector-space-axioms-x-y-be-x-y-for-R

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Question Number 1537 by 2closedStringsMeet last updated on 17/Aug/15 $S h o w b y u s e o f t h e c h a r a c t e r i s t i c s o f$ $a v e c t o r - s p a c e, t h a t t h e s e t \vec{x} \in R_{+}$ $with the following operations \oplus a n d$ $b u i l d s a v e c t o r - s p a c e .$ $* U s e v e c t o r - s p a c e a x i o m s$ $$\ast\overset{\rightarrow} {{x}}\oplus\overset{\rightarrow} {{y}}\:\:\:\:{be}\:\:\:\overset{\rightarrow} {{x}}\centerdot\overset{\rightarrow} {{y}}…

Category: Vector Calculus