(i) A force of magnitude 5 units acting parallel to $2\vec i - 2\vec j + \vec k$ displaces the point of application from $(1,2,3)$ to $(5,3,7)$. Find the work done by the force. (ii) If $A(-1,4,-3)$ is one end of a diameter AB of the sphere $x^2+y^2+z^2-3x-2y+2z-15=0$, then find the coordinates of B.

(i) A force of magnitude 5 units acting parallel t – Class 12 Mathematics

Question 106 of 162

2\vec i - 2\vec j + \vec k

(ii) If

A(-1,4,-3)

is one end of a diameter AB of the sphere

x^2+y^2+z^2-3x-2y+2z-15=0

, then find the coordinates of B.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 6mImportance★★★★★

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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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(i) Build the force vector from its magnitude and direction, then dot it with the displacement vector. (ii) Find the sphere's centre by completing the square, then use the midpoint formula for a diameter.

Part (i): Work done by the force

1. Unit vector along $2\vec i-2\vec j+\vec k$ .

$|2\vec i-2\vec j+\vec k|=\sqrt{4+4+1}=3\ \Rightarrow\ \hat u=\frac{2\vec i-2\vec j+\vec k}{3}$

2. Force vector of magnitude 5.

$\vec F=5\hat u=\frac{10}{3}\vec i-\frac{10}{3}\vec j+\frac53\vec k$

3. Displacement vector.

$\vec d=(5-1)\vec i+(3-2)\vec j+(7-3)\vec k=4\vec i+\vec j+4\vec k$

4. Work done $=\vec F\cdot\vec d$ .

$W=\frac{10}{3}(4)+\left(-\frac{10}{3}\right)(1)+\frac53(4)=\frac{40}{3}-\frac{10}{3}+\frac{20}{3}=\frac{50}{3}\text{ units}$

Part (ii): The far end $B$ of the diameter

1. Find the centre of the sphere $x^2+y^2+z^2-3x-2y+2z-15=0$ by completing the square: …

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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.(i) A force of magnitude 5 units acting parallel to 2i⃗−2j⃗+k⃗2\vec i - 2\vec j + \vec k2i−2j​+k displaces the point of application from (1,2,3)(1,2,3)(1,2,3) to (5,3,7)(5,3,7)(5,3,7). Find the work done by the force.

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Q.(i) A force of magnitude 5 units acting parallel to $2\vec i - 2\vec j + \vec k$ displaces the point of application from $(1,2,3)$ to $(5,3,7)$ . Find the work done by the force.

Unlock everything free for 14 days