=3i+πj+ck,v=4i−j−k, where is a constant. (a) Compute ‖u‖, ‖v‖, and u×v for the given vector in R3. (b) Verify the Cauchy-Schwarz inequality for the given pair of vector.

Here u=3i+πj+ck, v=4i−j−k, where c is a constant. (a) ||u||=32+π2+c2=3+π2+c2 ||v||=42+1+1=18 and u⋅v=(3i+πj+ck)⋅(4i−j−k)=43−π−c. (b) |u⋅v|=|43−π−c|≤3+π2+c218=||u||||v||.

Explained: "=3i+πj+ck,v=4i−j−k, where is a constant. (a) Compute ‖u‖,..."

High School
Geometry
Trigonometry
Trigonometric Functions

chillywilly12a

Answered question

2021-02-02

$= \sqrt{3 i} + π j + c k, v = 4 i - j - k$ , where is a constant.
(a) Compute $‖ u ‖$ , $‖ v ‖$ , and $u \times v$ for the given vector in R3.
(b) Verify the Cauchy-Schwarz inequality for the given pair of vector.

Answer & Explanation

Elberte

Skilled2021-02-03Added 95 answers

Here $u = \sqrt{3 i} + π j + c k, v = 4 i - j - k$ , where c is a constant.

(a) $| | u | | = \sqrt{{\sqrt{3}}^{2} + π^{2} + c^{2} = \sqrt{3 + π^{2} + c^{2}}}$
$| | v | | = \sqrt{4^{2} + 1 + 1} = \sqrt{18}$
and
$u \cdot v = (\sqrt{3} i + π j + c k) \cdot (4 i - j - k) = 4 \sqrt{3} - π - c .$
(b) $| u \cdot v | = | 4 \sqrt{3} - π - c | \leq \sqrt{3 + π^{2} + c^{2}} \sqrt{18} = | | u | | | | v | | .$

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