+91-8076753736[email protected]
logo
logo
-->

Vector Algebra

Question
CBSEENMA12034019
Wired Faculty App

Show that vector straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top is equally inclined to the axes OX. OY and OZ. 

Solution

Let straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space space straight k with hat on top
 therefore space space space space space straight a with rightwards arrow on top space equals space square root of left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared end root space equals space square root of 1 plus 1 plus 1 end root space equals square root of 3 therefore space space space space straight a with hat on top space equals space fraction numerator straight a over denominator open vertical bar straight a with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 3 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space equals space fraction numerator 1 over denominator square root of 3 end fraction straight i with hat on top space plus space fraction numerator 1 over denominator square root of 3 end fraction straight j with hat on top space plus space fraction numerator 1 over denominator square root of 3 end fraction straight k with hat on top therefore space space space space direction space cosines space of space straight a with rightwards arrow on top space are space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space space cos space straight alpha space equals fraction numerator 1 over denominator square root of 3 end fraction comma space space cos space straight beta space equals space fraction numerator 1 over denominator square root of 3 end fraction comma space space cos space straight gamma space equals space fraction numerator 1 over denominator square root of 3 end fraction
where straight alpha comma space straight beta comma space straight gamma are angles made by vector with axes.
therefore space space space space cos space straight alpha space equals space cos space straight beta space equals space cos space straight gamma rightwards double arrow space space space space space space space space space straight alpha space equals straight beta space equals straight gamma therefore space space space space given space vector space is space equally space inclined space to space the space axes space OX comma space OY comma space OZ.

Some More Questions From Vector Algebra Chapter

Classify the following measures as scalars and vectors.
time period

Classify the following measures as scalars and vectors.
distance

Classify the following measures as scalars and vectors.
force

Classify the following measures as scalars and vectors.
velocity

Classify the following measures as scalars and vectors.
work done

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation