Question

If PxQ=0 and Q x R = 0 then , | PxR| is equal to
a) zero
b)PR sinθ
c)PR cosθ
d)PR tanθ

Answer 1

Explanation:

a) zero

is correct answer

Answer 2

Explanation:

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0This implies either;

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0This implies either;P=0, or,

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0This implies either;P=0, or,Q= 0, or

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0This implies either;P=0, or,Q= 0, orBoth P, Q= 0, or

We are given that P × Q = 0, and Q × P = 0. From vector algebra we know that magnitude of P× Q is| P × Q| = |P|×|Q|× Sin ß, where ß is the angle between P and Q.P × Q = 0 ==> |P ×Q | = 0; ==> |P|× |Q| × Sin ß =0This implies either;P=0, or,Q= 0, orBoth P, Q= 0, orAngle ß= 0° or 180° . That is vectors P and Q are parallel or anti-parallel to each other

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