Find the magnitude and the direction cosines of the torque about the point (2, 0, -1) of a force 2hati+hatj-ha

Question 143 of 162

(2, 0, -1)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2022Subjective· 3mImportance★★★★★

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Concept understanding — Applications of Dot and Cross Product

The dot product $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ and cross product $\vec a\times\vec b$ (magnitude $|\vec a||\vec b|\sin\theta$ , direction perpendicular to both) are not just formulas — placed carefully, they reprove classical geometry theorems and compute two real mechanical quantities.

Vector proofs in geometry and trigonometry. The standard technique: put a convenient point (a triangle's vertex, a circle's centre) at the origin, write every other point as a position vector, and express every needed segment as a difference of position vectors. Then:

Perpendicularity is proved by showing a dot product is $0$ . E.g. if $O$ is a circle's centre and $M$ is the midpoint of chord $AB$ (position vectors $\vec a,\vec b$ with $|\vec a|=|\vec b|=r$ ), then $\vec{OM}=\tfrac{\vec a+\vec b}2$ and $\vec{AB}=\vec b-\vec a$ , and $\vec{OM}\cdot\vec{AB}=\tfrac12(|\vec b|^2-|\vec a|^2)=0$ .
Equal lengths / rectangles come from expanding $|\vec p\pm\vec q|^2=|\vec p|^2\pm2\vec p\cdot\vec q+|\vec q|^2$ and comparing.
Areas come from $\text{area of a parallelogram}=|\vec p\times\vec q|$ for adjacent sides $\vec p,\vec q$ ; a triangle is half that, and a general quadrilateral with diagonals $\vec d_1,\vec d_2$ has area $\tfrac12|\vec d_1\times\vec d_2|$ (proved by splitting along one diagonal, since the two triangles on either side add — same-sense cross products — instead of subtracting).
Compound-angle identities ( $\cos(\alpha\mp\beta)$ , $\sin(\alpha\pm\beta)$ ) drop out of dotting or crossing two unit vectors $\hat i\cos\alpha+\hat j\sin\alpha$ and $\hat i\cos\beta\pm\hat j\sin\beta$ placed at angles $\alpha,\pm\beta$ to the $x$ -axis. …

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Forms the position vector from the given point to the origin (a point on the force's line of action), crosses it with the force to get the torque, then normalizes for the direction cosines.

Force: $\vec F=2\hat i+\hat j-\hat k$ , acting along a line through the origin.
Take $\vec r$ as the vector from the point $(2,0,-1)$ (about which the torque is measured) to any point on the line of action — use the origin: $\vec r=(0-2)\hat i+(0-0)\hat j+(0-(-1))\hat k=-2\hat i+0\hat j+1\hat k$ .
Torque $\vec\tau=\vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\-2&0&1\\2&1&-1\end{vmatrix}$ .
$\hat i$ -component: $0(-1)-1(1)=-1$ .
$\hat j$ -component: $-\big((-2)(-1)-1(2)\big)=-(2-2)=0$ .
$\hat k$ -component: $(-2)(1)-0(2)=-2$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2019Set ANNUAL1 markMCQ
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
(a) $\dfrac{5\pi}{3}$
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{6}$
(d) $\dfrac{2\pi}{3}$
›Reveal solutionSolution
From $\vec a+\vec b+\vec c=\vec 0$ with $|\vec a|=3,|\vec b|=4,|\vec c|=5$ , the angle between $\vec a$ and $\vec b$ works out to $\dfrac{\pi}{2}$ .

From $\vec a+\vec b+\vec c=\vec0$ , we get $\vec c=-(\vec a+\vec b)$ .

Take magnitudes squared: $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2$ .

Substitute the given magnitudes: $5^2=3^2+2\vec a\cdot\vec b+4^2$ , i.e. $25=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b$ .

So $2\vec a\cdot\vec b=0\Rightarrow \vec a\cdot\vec b=0$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\vec{PR} = 2\vec i + \vec j + \vec k$
(a) $5\sqrt3$
(b) $10\sqrt3$
(c) $\dfrac{5\sqrt3}{2}$
(d) $\dfrac{3}{2}$
›Reveal solutionSolution
The area of a quadrilateral in terms of its diagonal vectors $\vec d_1,\vec d_2$ is $\frac12|\vec d_1\times\vec d_2|$ ; here the diagonals are $\vec{PR}$ and $\vec{QS}$ , so compute their cross product and halve its magnitude.

For quadrilateral PQRS, the diagonals are $PR$ and $QS$ , and the standard vector-algebra result is $\text{Area}(PQRS) = \frac12\,|\vec{PR}\times\vec{QS}|.$

Given $\vec{PR}=2\vec i+\vec j+\vec k$ and $\vec{QS}=-\vec i+3\vec j+2\vec k$ .

Compute the cross product: $\vec{PR}\times\vec{QS}=\begin{vmatrix}\vec i & \vec j & \vec k\\ 2 & 1 & 1\\ -1 & 3 & 2\end{vmatrix}$

$\vec i$ -component: $(1)(2)-(1)(3) = 2-3=-1$ .

$\vec j$ -component: $-[(2)(2)-(1)(-1)] = -(4+1)=-5$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$\vec{i} - \vec{j}$
(a) $0$
(b) $1$
(c) $-1$
(d) $2$
›Reveal solutionSolution
The vector $\vec i-\vec j$ has zero component along $\vec k$ , so its projection on the $z$ -axis is $0$ .

The projection of a vector $\vec r=x\vec i+y\vec j+z\vec k$ on the $z$ -axis equals $\vec r\cdot\vec k=z$ .

Here $\vec r=\vec i-\vec j=1\vec i-1\vec j+0\vec k$ .

So $\vec r\cdot\vec k=0$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$3\vec i + \vec j - \vec k$
(a) $10\sqrt3$
(b) $6\sqrt{30}$
(c) $\dfrac{3}{2}\sqrt{30}$
(d) $3\sqrt{30}$
›Reveal solutionSolution
Recovering the second side from the diagonal and given side, then taking the cross product, gives an area of $3\sqrt{30}$ .

In a parallelogram with adjacent sides $\vec a$ and $\vec b$ from the same vertex, the diagonal from that vertex is $\vec d=\vec a+\vec b$ .

Given side $\vec a=\vec i-3\vec j+4\vec k$ and diagonal $\vec d=3\vec i+\vec j-\vec k$ , the other side is $\vec b=\vec d-\vec a=(3-1)\vec i+(1-(-3))\vec j+(-1-4)\vec k=2\vec i+4\vec j-5\vec k$ .

Area of the parallelogram $=|\vec a\times\vec b|$ .

Compute the cross product: $\vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\1&-3&4\\2&4&-5\end{vmatrix}$ $i$ -component: $(-3)(-5)-(4)(4)=15-16=-1$ . $j$ -component: $-[(1)(-5)-(4)(2)]=-[-5-8]=13$ . …
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Q.Find the magnitude and the direction cosines of the torque about the point (2,0,−1)(2, 0, -1)(2,0,−1) of a force 2i^+j^−k^2\hat{i}+\hat{j}-\hat{k}2i^+j^​−k^, whose line of action passes through the origin.

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Q.Find the magnitude and the direction cosines of the torque about the point $(2, 0, -1)$ of a force $2\hat{i}+\hat{j}-\hat{k}$ , whose line of action passes through the origin.

Unlock everything free for 14 days