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Please solve with a detailed explanation?

The velocity of a particle is $\to v = 6 ˆ i + 2 ˆ j - 2 ˆ k$ $vec v =6hati +2hatj-2hatk$ .The component of the velocity parallel to vector $\to a = ˆ i + ˆ j + ˆ k$ $veca=hati +hatj +hatk$ in vector form is

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Answer 1 · 2018-02-23T17:41:34

We can say , $\to v . \to a = ∣ ∣ \to v ∣ ∣ ∣ ∣ \to a ∣ ∣ cos θ$ $vec v.vec a = |vec v| |vec a| cos theta$ (where, $θ$ $theta$ is the angle between the two vector)

So, $∣ ∣ \to v ∣ ∣ cos θ = \frac{\to v . \to a}{∣ ∣ \to a ∣ ∣}$ $|vec v| cos theta = (vec v . vec a)/(|vec a|)$

Now, $∣ ∣ \to v ∣ ∣ cos θ$ $|vec v| cos theta$ is defined as the projection of $\to v$ $vec v$ on $\to a$ $vec a$

So, the magnitude of the projection is $\frac{6 + 2 - 2}{\sqrt{3}} = 2 \sqrt{3}$ $(6+2-2)/sqrt(3) = 2sqrt(3)$ (given , $\to v = 6 i + 2 j - 2 k$ $vec v=6i+2j-2k$ and $\to a = i + j + k$ $vec a=i+j+k$ )

So,the vector will be $2 \sqrt{3} \frac{i + j + k}{| (i + j + k) |} = 2 (i + j + k)$ $2sqrt(3)(i+j+k)/(|(i+j+k)| )= 2(i+j+k)$ (as parallel to $\to a$ $vec a$ ,so just multiplied the unit vector along $\to a$ $vec a$ with the magnitude of projection to get the vector)

Answer 2 · 2018-02-23T17:44:56

$2 ˆ i + 2 ˆ j + 2 ˆ k$ $2hat{i}+2hat{j}+2hat{k}$

Explanation:

Given a vector $\to v$ $vec{v}$ and a unit vector $ˆ n$ $hat{n}$ , you can divide the vector $\to v$ $vec{v}$ into two parts

$\to v = \to p + \to q$ $vec{v} = vec{p}+vec{q}$

where $\to p$ $vec{p}$ and $\to q$ $vec{q}$ are parallel and perpendicular to $ˆ n$ $hat{n}$ , respectively. Now, the magnitude of $\to p$ $vec{p}$ is given by $\to p \cdot ˆ n$ $vec{p} cdot hat{n}$ , and since $\to q \cdot ˆ n = 0$ $vec{q} cdot hat{n}=0$ we have

$∣ ∣ \to p ∣ ∣ = \to p \cdot ˆ n = \to v \cdot ˆ n$ $|vec{p}| = vec{p} cdot hat {n} = vec{v} cdot hat{n}$

Thus,

$\to p = ∣ ∣ \to p ∣ ∣ ˆ n = (\to v \cdot ˆ n) ˆ n$ $vec{p} = |vec{p}| hat{n} = ( vec{v} cdot hat{n}) hat{n}$

Since the unit vector in the direction of $\to a$ $vec{a}$ is given by $\frac{\to a}{∣ ∣ \to a ∣ ∣}$ ${vec{a}}/{|vec{a}|}$ , the component of $\to v$ $vec{v}$ parallel to $\to a$ $vec{a}$ is given by

$⎛ ⎜ ⎝ \to v \cdot \frac{\to a}{∣ ∣ \to a ∣ ∣} ⎞ ⎟ ⎠ \frac{\to a}{∣ ∣ \to a ∣ ∣} = \frac{(\to v \cdot \to a) \to a}{{∣ ∣ \to a ∣ ∣}^{2}}$ $(vec{v} cdot {vec{a}}/{|vec{a}|}){vec{a}}/{|vec{a}|}= {(vec{v} cdot vec{a})vec{a}}/{|vec{a}|^2$

So, for the given problem, the component that we seek is

$\frac{(6 ˆ i + 2 ˆ j - 2 ˆ k) \cdot (ˆ i + ˆ j + ˆ k)}{(ˆ i + ˆ j + ˆ k) \cdot (ˆ i + ˆ j + ˆ k)} (ˆ i + ˆ j + ˆ k) = 2 ˆ i + 2 ˆ j + 2 ˆ k$ ${(6 hat{i}+2 hat{j}-2 hat{k}) cdot (hat{i}+hat{j}+hat{k})} /{(hat{i}+hat{j}+hat{k})cdot (hat{i}+hat{j}+hat{k})} (hat{i}+hat{j}+hat{k}) = 2hat{i}+2hat{j}+2hat{k}$

Answer 3 · 2018-02-23T18:03:06

Component of the velocity parallel to the vector
$ˆ i + ˆ j + ˆ k$ $hati+hatj+hatk$
is

$2 (ˆ i + ˆ j + ˆ k)$ $2(hati+hatj+hatk)$

Explanation:

Given:

$\to v = 6 ˆ i + 2 ˆ j - 2 ˆ k$ $vecv=6hati+2hatj-2hatk$

Projection needed along

$\to a = ˆ i + ˆ j + ˆ k$ $veca=hati+hatj+hatk$

$\frac{(\to v \cdot \to a)}{{∣ ∣ \to a ∣ ∣}^{2}} \to a$ $((vecv*veca))/(|veca|^2)veca$

$\to v \cdot \to a = (6 ˆ i + 2 ˆ j - 2 ˆ k) \cdot (ˆ i + ˆ j + ˆ k)$ $vecv*veca=(6hati+2hatj-2hatk)*(hati+hatj+hatk)$

$= 6 \times 1 + 2 \times 1 - 2 \times 1 = 6 + 2 - 2 = 6$ $=6xx1+2xx1-2xx1=6+2-2=6$

$\to v \cdot \to a = 6$ $vecv*veca=6$

${∣ ∣ \to a ∣ ∣}^{2} = {∣ ∣ ˆ i + ˆ j + ˆ k ∣ ∣}^{2} = 1^{2} + 1^{2} + 1^{2} = 1 + 1 + 1 = 3$ $|veca|^2=|hati+hatj+hatk|^2=1^2+1^2+1^2=1+1+1=3$

${∣ ∣ \to a ∣ ∣}^{2} = 3$ $|veca|^2=3$

$\frac{(\to v \cdot \to a)}{{∣ ∣ \to a ∣ ∣}^{2}} \to a = \frac{6}{3} (ˆ i + ˆ j + ˆ k) = 2 (ˆ i + ˆ j + ˆ k$ $((vecv*veca))/(|veca|^2)veca=6/3(hati+hatj+hatk)=2(hati+hatj+hatk$

$\frac{(\to v \cdot \to a)}{{∣ ∣ \to a ∣ ∣}^{2}} \to a = 2 (ˆ i + ˆ j + ˆ k)$ $((vecv*veca))/(|veca|^2)veca=2(hati+hatj+hatk)$

Component of the velocity parallel to the vector
$ˆ i + ˆ j + ˆ k)$ $hati+hatj+hatk)$
is

$2 (ˆ i + ˆ j + ˆ k)$ $2(hati+hatj+hatk)$

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The velocity of a particle is $\to v = 6 ˆ i + 2 ˆ j - 2 ˆ k$ $vec v =6hati +2hatj-2hatk$ .The component of the velocity parallel to vector $\to a = ˆ i + ˆ j + ˆ k$ $veca=hati +hatj +hatk$ in vector form is

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The velocity of a particle is vec v =6hati +2hatj-2hatk→v=6ˆi+2ˆj−2ˆk vec v =6hati +2hatj-2hatk.The component of the velocity parallel to vector veca=hati +hatj +hatk→a=ˆi+ˆj+ˆkveca=hati +hatj +hatk in vector form is

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The velocity of a particle is $\to v = 6 ˆ i + 2 ˆ j - 2 ˆ k$ $vec v =6hati +2hatj-2hatk$ .The component of the velocity parallel to vector $\to a = ˆ i + ˆ j + ˆ k$ $veca=hati +hatj +hatk$ in vector form is