Q20 of 43 Page 1

Let and

Find a vector which is perpendicular to both and and

Let,


Given, ⟘


∴ &


Hence,



⇒ x + 4y + 2z = 0 …(1)



⇒ 3x – 2y + 7z = 0 …(2)


Solving equation 1 and 2 using Cramer's rule, taking z as an independent variable, we have -



⇒ (Say)


∴ x = 32λ, y = -λ, z = -14λ


Also, = 18


∴


⇒ 2x – y + 4z = 18


⇒ 2 (32λ) – (-λ) + 4 (-14λ) = 18


⇒ 9λ = 18


∴ λ = 2


Hence, x = 64, y = -2 & z = -28


Or


The required vector is .


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