The position vectors of points A-B-C and D are A- 3- 4- 5k- B- 4- 5- 6k- C - 7- 9- 3k- and D - 4- 6 - then the displacement vectors AB and CD are

A = 3î+ 4ĵ+5 k̂ B = 4 î+ 5ĵ+ 6 k̂ C = 7î+ 9ĵ+3 k̂ D = 4î+ 6ĵ AB = B A = (4î+ 5ĵ+ 5k̂) (3î+ 4ĵ+ 5k̂) = î+ ĵ+ k̂ nbsp; CD = D C =(4î+5ĵ) (7î+ 9 ĵ+ 3k̂) ...

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Byju's Answer

Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

The position ...

Question

The position vectors of points $A, B, C$ and $D$ are $A = 3^i + 4^j + 5^k$ , $B = 4^i + 5^j + 6^k, C = 7^i + 9^j + 3^k$ and $D = 4^i + 6^j$ , then the displacement vectors $- - \to A B$ and $- - \to C D$ are

Perpendicular

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Parallel

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Antiparallel

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Inclined at an angle of

60^{\circ}

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Solution

The correct option is C Antiparallel
$\to A = 3^i + 4^j + 5^k$
$\to B = 4^i + 5^j + 6^k$
$\to C = 7^i + 9^j + 3^k$
$\to D = 4^i + 6^j$
$- - \to A B = \to B - \to A$
$= (4^i + 5^j + 5^k) - (3^i + 4^j + 5^k)$
$=^i +^j +^k$

$- - \to C D = \to D - \to C$
$= (4^i + 5^j) - (7^i + 9^j + 3^k)$
$= - 3^i - 3^j - 3^k$
SInce $- - \to C D = λ (- - \to A B) (where, λ = - 3)$
Therefore , they are antiparallel.

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The position vectors of points A,B,C and D are A=3^i+4^j+5^k, B=4^i+5^j+6^k,C=7^i+9^j+3^k and D=4^i+6^j, then the displacement vectors −−→AB and −−→CD are

The position vectors of points $A, B, C$ and $D$ are $A = 3^i + 4^j + 5^k$ , $B = 4^i + 5^j + 6^k, C = 7^i + 9^j + 3^k$ and $D = 4^i + 6^j$ , then the displacement vectors $- - \to A B$ and $- - \to C D$ are