A unit vector perpendicular to the plane of a = 2 ^ i − 6 ^ j − 3 ^ k , b = 4 ^ i + 3 ^ j − ^ k is Options: A . &nbsp 4 ^ i + 3 ^ j − ^ k √ 26 B . &nbsp 2 ^ i − 6 ^ j − 3 ^ k 7 C . &nbsp 3 ^ i − 2 ^ j + 6 ^ k 7 D . &nbsp 2 ^ i − 3 ^ j − 6 ^ k 7

Answer: Option C : C a × b = ∣ ∣ ∣ ∣ ∣ ^ i ^ j ^ k 2 − 6 − 3 4 3 − 1 ∣ ∣ ∣ ∣ ∣ = ^ i ( 6 + 9 ) − ^ j ( − 2 + 12 ) + ^ k ( 6 + 24 ) = 15 ^ i − 10 ^ j + 30 ^ k | a × b | = √ 225 + 100 + 900 = 35 Unit vector normal to the plane = 15 ^ i − 10 ^ j + 30 ^ k 35 = 3 ^ i − 2 ^ j + 6 ^ k 7

A unit vector perpendicular to the plane of a=2^i−6^j−3^k,b=4^i+

Discussion Forum : Vector Algebra

Question -

A unit vector perpendicular to the plane of $a = 2^i - 6^j - 3^k, b = 4^i + 3^j -^k$ is

Options:

A .

\frac{4^i + 3^j -^k}{\sqrt{26}}

B .

\frac{2^i - 6^j - 3^k}{7}

C .

\frac{3^i - 2^j + 6^k}{7}

D .

\frac{2^i - 3^j - 6^k}{7}

Answer: Option C
:
C

a \times b = ∣ ∣∣∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 6 & - 3 4 & 3 & - 1 \end{matrix} ∣ ∣∣∣ ∣ =^i (6 + 9) -^j (- 2 + 12) +^k (6 + 24) = 15^i - 10^j + 30^k

| a \times b | = \sqrt{225 + 100 + 900} = 35

Unit vector normal to the plane =

\frac{15^i - 10^j + 30^k}{35} = \frac{3^i - 2^j + 6^k}{7}

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Discussion Forum : Vector Algebra

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