(i) Show that the points whose position vectors are $4\vec{i} - 3\vec{j} + \vec{k}$, $2\vec{i} - 4\vec{j} + 5\vec{k}$, $\vec{i} - \vec{j}$ form a right angled triangle. (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 co-ordinate of B.

> [!TLDR] > (i) Use vector differences and the dot product / Pythagoras to locate the right angle. > (ii) Use the sphere's general-equation centre formula together with the midpoint-of-diameter property to find B. > > 1. Position vectors: $\vec{OA}=4\vec i-3\vec j+\vec k$, $\vec{OB}=2\vec i-4\vec j+5\vec k$, $\vec{OC}=\vec i-\vec j$ (naming the three given points $A,B,C$). > 2. $\vec{AB}=\vec{OB}-\vec{OA}=(2-4)\vec i+(-4+3)\vec j+(5-1)\vec k=-2\vec i-\vec j+4\vec k$, so $|\vec{AB}|^2=4+1+16=21$. > 3. $\vec{AC}=\vec{OC}-\vec{OA}=(1-4)\vec i+(-1+3)\vec j+(0-1)\vec k=-3\vec i+2\vec j-\vec k$, so $|\vec{AC}|^2=9+4+1=14$. > 4. $\vec{BC}=\vec{OC}-\vec{OB}=(1-2)\vec i+(-1+4)\vec j+(0-5)\vec k=-\vec i+3\vec j-5\vec k$, so $|\vec{BC}|^2=1+9+25=35$. > 5. $\vec{AB}\cdot\vec{AC}=(-2)(-3)+(-1)(2)+(4)(-1)=6-2-4=0$, so $\vec{AB}\perp\vec{AC}$, i.e. the angle at $A$ is $90^\circ$. > 6. Cross-check with Pythagoras: $|AB|^2+|AC|^2=21+14=35=|BC|^2$ — consistent, so $\triangle ABC$ is right-angled at $A$. …

(i) Show that the points whose position vectors ar – Class 12 Mathematics

Question 125 of 162

4\vec{i} - 3\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 co-ordinate of B.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★

77% · 125/162 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Use vector differences and the dot product / Pythagoras to locate the right angle.
Use the sphere's general-equation centre formula together with the midpoint-of-diameter property to find B.

Position vectors: $\vec{OA}=4\vec i-3\vec j+\vec k$ , $\vec{OB}=2\vec i-4\vec j+5\vec k$ , $\vec{OC}=\vec i-\vec j$ (naming the three given points $A,B,C$ ).
$\vec{AB}=\vec{OB}-\vec{OA}=(2-4)\vec i+(-4+3)\vec j+(5-1)\vec k=-2\vec i-\vec j+4\vec k$ , so $|\vec{AB}|^2=4+1+16=21$ .
$\vec{AC}=\vec{OC}-\vec{OA}=(1-4)\vec i+(-1+3)\vec j+(0-1)\vec k=-3\vec i+2\vec j-\vec k$ , so $|\vec{AC}|^2=9+4+1=14$ .
$\vec{BC}=\vec{OC}-\vec{OB}=(1-2)\vec i+(-1+4)\vec j+(0-5)\vec k=-\vec i+3\vec j-5\vec k$ , so $|\vec{BC}|^2=1+9+25=35$ .
$\vec{AB}\cdot\vec{AC}=(-2)(-3)+(-1)(2)+(4)(-1)=6-2-4=0$ , so $\vec{AB}\perp\vec{AC}$ , i.e. the angle at $A$ is $90^\circ$ .
Cross-check with Pythagoras: $|AB|^2+|AC|^2=21+14=35=|BC|^2$ — consistent, so $\triangle ABC$ is right-angled at $A$ . …

Start your 14-day free trial to unlock the full solution →

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 $2\vec a\cdot\vec b=0\Rightarrow \vec a\cdot\vec b=0$ . …
Start your 14-day free trial to see the full solution →
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: $-[(2)(2)-(1)(-1)] = -(4+1)=-5$ . …
Start your 14-day free trial to see the full solution →
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 $\vec r\cdot\vec k=0$ . …
Start your 14-day free trial to see the full solution →
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: $-[(1)(-5)-(4)(2)]=-[-5-8]=13$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ26 — The point of intersection of the lines vecr = (-veci + 2vecj + 3veck) …→Next questionQ28 — If the foot of the perpendicular drawn from the origin to the plane is…

Q.(i) Show that the points whose position vectors are 4i⃗−3j⃗+k⃗4\vec{i} - 3\vec{j} + \vec{k}4i−3j​+k, 2i⃗−4j⃗+5k⃗2\vec{i} - 4\vec{j} + 5\vec{k}2i−4j​+5k, i⃗−j⃗\vec{i} - \vec{j}i−j​ form a right angled triangle.

🎓Unlock everything free for 14 days

Q.(i) Show that the points whose position vectors are $4\vec{i} - 3\vec{j} + \vec{k}$ , $2\vec{i} - 4\vec{j} + 5\vec{k}$ , $\vec{i} - \vec{j}$ form a right angled triangle.

Unlock everything free for 14 days