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3-1.Vectors
easy
$ (4, -4, 0)$ અને $(-2,- 2, 0)$ બિંદુ વચ્ચે રહેલ સદીશનું મૂલ્ય કેટલું થાય?
A
$6$
B
$5\sqrt 2 $
C
$4$
D
$2\sqrt {10} $
Solution
(d)$\vec r = {\vec r_2} – {\vec r_1} = ( – 2\hat i – 2\hat j + 0\hat k) – (4\hat i – 4\hat j + 0\hat k)$
$⇒$ $\vec r = – 6\hat i + 2\hat j + 0\hat k$
$\therefore |\vec r| = \sqrt {{{( – 6)}^2} + {{(2)}^2} + {0^2}} = \sqrt {36 + 4} = \sqrt {40} = 2\sqrt {10} $
Standard 11
Physics
