Find a unit vector perpendicular to each of the vector a- - b- and a-b- where a- - 3- - 2- - 2k- and b- - - - 2- - 2k-

Given : a⃗ = 3î + 2ĵ + 2k̂ and b⃗ = î + 2ĵ 2k̂ (a⃗+b⃗) = (3î+2ĵ+2k̂)+(î+2ĵ 2k̂) ⇒ (a⃗ + b⃗) = 4î + 4ĵ + 0k̂ (a⃗ b⃗) = (3î + 2ĵ+2k̂) (î+2ĵ 2k̂) ⇒ (a⃗ b⃗)=2 ...

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Byju's Answer

Standard VII

History

Early Kakatiyas and Rudradeva

Find a unit v...

Question

Find a unit vector perpendicular to each of the vector $\to a + \to b and \to a - \to b, where \to a = 3^i + 2^j + 2^k and \to b =^i + 2^j - 2^k$

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Solution

Given : $\to a = 3^i + 2^j + 2^k and \to b =^i + 2^j - 2^k$

$(\to a + \to b) = (3^i + 2^j + 2^k) + (^i + 2^j - 2^k)$

$\Rightarrow (\to a + \to b) = 4^i + 4^j + 0^k$

$(\to a - \to b) = (3^i + 2^j + 2^k) - (^i + 2^j - 2^k)$

$\Rightarrow (\to a - \to b) = 2^i + 0^j + 4^k$

$Let \to c = (\to a + \to b) \times (\to a - \to b)$

$\to c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 4 & 4 & 0 2 & 0 & 4 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \to c =^i [(4 \times 4) - (0 \times 0)]$

$-^j [(4 \times 4) - (2 \times 0)]$

$+^k [(4 \times 0) - (2 \times 4)]$

$\Rightarrow \to c =^i (16 - 0) -^j (16 - 0) +^k (0 - 8)$

$\Rightarrow \to c = 16^i - 16^j - 8^k$

$Unit vector \to c = \frac{\to c}{magnitude of \to c}$

$\to c = \frac{(16^i - 16^j - 8^k)}{\sqrt{(16)^{2} + (- 16)^{2} + (- 8)^{2}}}$

$\to c = \frac{(16^i - 16^j - 8^k)}{\sqrt{256 + 256 + 64}}$

$\to c = \frac{(16^i - 16^j - 8^k)}{\sqrt{576}}$

$\to c = \frac{(16^i - 16^j - 8^k)}{24}$

$\to c = \frac{2}{3}^i - \frac{2}{3}^j - \frac{1}{3}^k$

$∴ Unit vector \to c (^c) = \frac{2}{3}^i - \frac{2}{3}^j - \frac{1}{3}^k$

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Early Kakatiyas and Rudradeva

Standard VII History

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Find a unit vector perpendicular to each of the vector →a+→b and →a−→b,where →a=3^i+2^j+2^k and →b=^i+2^j−2^k

Find a unit vector perpendicular to each of the vector $\to a + \to b and \to a - \to b, where \to a = 3^i + 2^j + 2^k and \to b =^i + 2^j - 2^k$