8. Suppose I is a triangle formed by placing three points lircle, two of which lie on the circle's diameter. Ube on a the previous problem to show I is a right triangle.
8. Suppose I is a triangle formed by placing three points lircle, two of which lie on the circle's diameter. Ube on a the previous problem to show I is a right triangle.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question

Transcribed Image Text:8. Suppose I is a triangle formed by placing three points
circle, two of which lie on the circle's diameter. Use
the previous problem to show I is a right triangle.
on a
XXX Xper 3-4 nitiat

Transcribed Image Text:congle.
Given that V₁W € Rn and 1||v|| = ||w|l.
We have to show that vtw and v-w are orthogonal
*Orthogonal vectors: Let
V
be
an inner product space and
orthogonal if (UV ro
U₁vbe vectors in V. Then vand v are
* properties: Let V be an inner product space them for all
virwEv
hove
ille
(i) (ut V₁06) = (v₁ ) + (1₁0)
(ii) (v₁v-w) = (U₁V) - (U₁0)
*If v be an real inner product space, then (U,V) = (V₁U) for
all v₁V E Ve
* Norm of a vector: Letv be an inner product space and
VEV. Then norm of v, denoted by IVII, is defined as 11V1= 5
Note: As per our quidelines. we are supposed to solve the STVU)
that
EIR and 11v11 = ||w|l.
first
one
only,
Hene
given
v, w
Now, (V+∞, v-w) = (v₁ V-) + (W₁V-W)
= (v₁v) - (V₁w) + (w, v) - (W₁w)
= || v11 ² - (V₁W) + (v₁w) - || || ²
2
= || v || ²_ ||w| || ²
2
= || V || ² - Il vii"
real inner
is al
product space, (WV) = (v₁)
And 11 vil = √(V₁V).
i.e. (V+W₁v-w)=0XE
=J Vtw and v- ore
Hence, vtw and v-c are
IR
[: || | || = |||||
orthogonal.
n
orthogonal
~(2)9
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