Two forces F-1-5-10-20k- and F-2-10-5-20 k- act on a single point- The angle between F-1- and F-2- is nearly

Cos θ =F⃗_⃗1⃗.F⃗_⃗2⃗/|F⃗_⃗1⃗||F_2| =(5î+10ĵ 20k̂).(10î 5ĵ 15k̂)/√(25+100+400)√(100+25+225)=50 50+300/√(525)√(350) ⇒ cos θ =1/√(2)∴θ =45^∘ ...

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Byju's Answer

Standard XIII

Physics

Addition and Subtraction in Unit Vector Notation

Two forces F⃗...

Question

Two forces $\vec{F_1}=5\hat{i}+10\hat{j}-20\hat{k}$ and $_{2} = 10^i - 5^j - 20^k$ act on a single point. The angle between $_{1}$ and $_{2}$ is nearly

$30^{\circ}$

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$45^{\circ}$

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$60^{\circ}$

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$90^{\circ}$

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Solution

The correct option is B
$45^{\circ}$

$c o s θ = \frac{_{1} ._{2}}{|_{1} | | F_{2} |} = \frac{(5^i + 10^j - 20^k) . (10^i - 5^j - 15^k)}{\sqrt{25 + 100 + 400} \sqrt{100 + 25 + 225}} = \frac{50 - 50 + 300}{\sqrt{525} \sqrt{350}} \Rightarrow c o s θ = \frac{1}{\sqrt{2}} ∴ θ = 45^{\circ}$

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Two forces $\vec{F_1}=5\hat{i}+10\hat{j}-20\hat{k}$ and $_{2} = 10^i - 5^j - 20^k$ act on a single point. The angle between $_{1}$ and $_{2}$ is nearly

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and between

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Under these conditions, forces cannot remain in equilibrium. Reason: No angle is greater than

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Two forces \(\vec{F_1}=5\hat{i}+10\hat{j}-20\hat{k}\) and →F2=10^i−5^j−20^k act on a single point. The angle between →F1 and →F2 is nearly

The correct option is B 45∘ cosθ=→F1.→F2|→F1||F2|=(5^i+10^j−20^k).(10^i−5^j−15^k)√25+100+400√100+25+225=50−50+300√525√350⇒cosθ=1√2∴θ=45∘

Two forces \(\vec{F_1}=5\hat{i}+10\hat{j}-20\hat{k}\) and $_{2} = 10^i - 5^j - 20^k$ act on a single point. The angle between $_{1}$ and $_{2}$ is nearly

The correct option is B
$45^{\circ}$

$c o s θ = \frac{_{1} ._{2}}{|_{1} | | F_{2} |} = \frac{(5^i + 10^j - 20^k) . (10^i - 5^j - 15^k)}{\sqrt{25 + 100 + 400} \sqrt{100 + 25 + 225}} = \frac{50 - 50 + 300}{\sqrt{525} \sqrt{350}} \Rightarrow c o s θ = \frac{1}{\sqrt{2}} ∴ θ = 45^{\circ}$