Find unit vector of a given vector a- - 2- - 3- - 4k-

a⃗ = 2î 3ĵ + 4k̂ Magnitude of a⃗ : √(2^2+( 3)^2+4^2) = √(29) Unit vector: 2√(29)î 3√(29)ĵ + 4√(29)k̂ ...

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Standard XII

Physics

Unit Vectors

Find unit vec...

Question

Find unit vector of a given vector $\to a = 2^i - 3^j + 4^k$

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Solution

$\to a = 2^i - 3^j + 4^k$

Magnitude of $\to a$ : $\sqrt{2^{2} + (- 3)^{2} + 4^{2}} = \sqrt{29}$
Unit vector: $\frac{2}{\sqrt{29}}^i - \frac{3}{\sqrt{29}}^j + \frac{4}{\sqrt{29}}^k$

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Q. A unit vector perpendicular to

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If $\to a$ and $\to b$ are vectors such that $| \to a + \to b | = \sqrt{29}$ and $\to a \times (2^i + 3^j + 4^k) = (2^i + 3^j + 4^k) \times \to b,$ then a possible value of $(\to a + \to b) . (- 7^i + 2^j + 3^k)$ is

Q. If

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Find unit vector of a given vector →a=2^i−3^j+4^k

→a=2^i−3^j+4^kMagnitude of →a: √22+(−3)2+42=√29Unit vector: 2√29^i−3√29^j+4√29^k

Find unit vector of a given vector $\to a = 2^i - 3^j + 4^k$

$\to a = 2^i - 3^j + 4^k$

Magnitude of $\to a$ : $\sqrt{2^{2} + (- 3)^{2} + 4^{2}} = \sqrt{29}$
Unit vector: $\frac{2}{\sqrt{29}}^i - \frac{3}{\sqrt{29}}^j + \frac{4}{\sqrt{29}}^k$