5 Marks Questions

Question 15 Marks

If the vectors $\big(\sec^2\text{A}\big)\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\hat{\text{i}}+\big(\sec^2\text{B}\big)+\hat{\text{k}},\hat{\text{i}}+\hat{\text{j}}+\big(\sec^2\text{C}\big)\hat{\text{k}}$ are coplanar, then find the value of $\text{cosec}^2\text{A}+\text{cosec}^2\text{B}+\text{cosec}^2\text{C}.$

Answer

Let: $\vec{\text{a}}=\big(\sec^2\text{A}\big)\hat{\text{i}}+\hat{\text{j}},\vec{\text{b}}=\hat{\text{i}}+(\sec^2\text{B})\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+(\sec^2\text{C})\hat{\text{k}}$
We know that three vectors are coplanar if their scaler triple product is zero i.e., $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
Here, $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\begin{vmatrix}\sec^2\text{A}&1&1\\1&\sec^2\text{B}&1\\1&1&\sec^2\text{C} \end{vmatrix}=0$
$\Rightarrow\sec^2\text{A}\big[\big(\sec^2\text{B}\times\sec^2\text{C}\big)\\-1\big]-1\big(\sec^2\text{C}-1\big)+1\big(1-\sec^2\text{B}\big)=0$
$\Rightarrow\sec^2\text{A}\sec^2\text{B}\sec^2\text{C}-\sec^2\text{A}-\sec^2\text{C}+1+1-\sec^2\text{B}=0$
$\Rightarrow\big(1+\tan^2\text{A}\big)\big(1+\tan^2\text{B}\big)\big(1+\tan^2\text{C}\big)\\-\big(1+\tan^2\text{A}\big)-\big(1+\tan^2\text{C}\big)+1=1-\big(1+\tan^2\text{B}\big)=0$
$\Rightarrow1+\tan^2\text{A}+\tan^2\text{B}+\tan^2\text{C}+\tan^2\text{A}\tan^2\text{B}\\+\tan^\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}+\tan^2\text{A}\tan^2\text{B}\tan^\text{C}1\\-\tan^2\text{A}-1-\tan^2\text{C}$
$\tan^2\text{A}\tan^2\text{B}+\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}\\+\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}=0$
$\Rightarrow\tan^2\text{A}\tan^2\text{B}+\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}\\=-\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}$
$\Rightarrow\frac{\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}}{\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}}=-1$
$\Rightarrow\cot^2\text{C}+\cot^2\text{A}+\cot^2\text{B}=-1$
$\Rightarrow\text{cosec}^2\text{C}-1+\text{cosec}^2\text{A}-1+\text{cosec}^2\text{B}-1=-1$
$\therefore\text{cosec}^2\text{A}+\text{cosec}^2\text{B}+\text{cosec}^2\text{C}=2$

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Question 25 Marks

Prove that: $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\big\}=0$

Answer

We have
$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}-\vec{\text{c}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$\big(\therefore\vec{\text{c}}\times\vec{\text{c}}=0\big)$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{a}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{b}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive 1)
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{a}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{a}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+0+0-0-0-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]$
$\big(\therefore\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]=\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]\big)$
$=0$

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Question 35 Marks

Find $\lambda$ for which the points A(3, 2, 1), B(4, $\lambda$, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar.

Answer

The points A, B, C and D will be coplanar if will be coplanar of any one of the following traces of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}},$ etc.
It is given that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
Thus, their scaler triple product $\big[\vec{\text{AB }}\vec{\text{AC }}\vec{\text{AD}}\big]$ is equal to zero.
Now,
Direction ratios of the $\vec{\text{PQ}}$ =(Direction ratios of vector Q) - (Direction ratios of the vector P)
Direction ratios of vector $\vec{\text{AB}}=(4-3),(\lambda-2),(5-1),\text{i. e. 1},\lambda, -2, 4$
Direction ratios of vector $\vec{\text{AC}}=(4-3),(2-2),(-2 -1),\text{i. e. } 3, 3, -2$
Direction ratios of vector $\vec{\text{AD}}=(6-3),(5-2),(-1-1),\text{i. e}. 3, 3, -2$
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=\begin{vmatrix}1&\lambda-2&4\\1&0&-3\\3&3&-2\end{vmatrix}$
$=1[0-(-9)]-(\lambda-2)[-2-(-9)]+4(3-0)=0$
$\Rightarrow7\lambda=35$
$\Rightarrow\lambda=5$

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Question 45 Marks

Find the value of $\lambda$ for which the four points with position vectors
$-\hat{\text{j}}-\hat{\text{k}},4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}},3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$ and $-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$ are co planar.

Answer

Let
position vector of $\text{A}=-\hat{\text{j}}-\hat{\text{k}}$
position vector of $\text{B}=4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}}$
position vector of $\text{C}=3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$
position vector of $\text{D}=-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
The four points are coplanar if the vectors $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
$\vec{\text{AB}}=4\hat{\text{i}}+6\hat{\text{j}}+(\lambda+1)\hat{\text{k}}$
$\vec{\text{AC}}=3\hat{\text{i}}+10\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{AD}}=4\hat{\text{i}}+5\hat{\text{j}}+5\hat{\text{k}}$
$\begin{vmatrix}4&6&(\lambda+1)\\3&10&5\\-4&5&5 \end{vmatrix}=0$
$4(50-25)-6(15+20)+(\lambda+1)(15+40)=0$
$100-210+55+55\lambda=0$
$55\lambda=55$
$\lambda=1$

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Question 55 Marks

Show that four points whose position vectors are
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{i}}-6\hat{\text{k}},2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.

Answer

DISCLAIMER: Given points are not coplaner.
Let A, B, C, D be the given points. The given points will be coplanar iff any one of the follewing triads of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}}$ etc.
In order to show that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar, we will have to show that their scaler triple
product i.e. $\Big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\Big]=0$
Using, $\vec{\text{PQ}}$ = Position vector of Q - position vector of P, we obtain
Now,
$\vec{\text{AB}}=(16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{AC}}=(3\hat{\text{i}}-6\hat{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=-3\hat{\text{i}}+7\hat{\text{j}}-6\hat{\text{k}}$
and, $\vec{\text{AD}}=(2\hat{\text{i}}-5\hat{\text{j}}+10\vec{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}$
$\therefore\Big[\vec{\text{AC}}\vec{\text{ AC }}\vec{\text{AD}}\Big]=\begin{vmatrix}10&-12&-4\\-3&7&-6\\-4&2&10 \end{vmatrix}$
$=10(70+12)+12(-30-24)-4(-6+28)=84$
Thus, the given points are not coplanar.

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Question 65 Marks

$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are the position vectors of points A, B and C respectively, prove that:
$\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is a vector perpendicular to the plane of triangle ABC.

Answer

We know that if any vector is perpendicular to all three sides of $\triangle\text{ABC},$ it must be perpendicular to the plane of $\triangle \text{ABC}.$
Now,
$\vec{\text{AB}}=\vec{\text{b}}-\vec{\text{a}},\vec{\text{BC}}=\vec{\text{c}}-\vec{\text{b}},\vec{\text{CA}}=\vec{\text{a}}-\vec{\text{c}}$
$\big($position vectors of A, B and C are $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}\big)$
We have
$\vec{\text{AB}}.\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{b}}-\vec{\text{a}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{b}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{a}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{a}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=0+0+\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-0-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-0$
$=0\ \big(\therefore\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]\big)$
$\vec{\text{BC}}\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{c}}-\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{c}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{c}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{b}}\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{c}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{c}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+0+0-0-0-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=0$ $=0 \ \big(\therefore\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}=\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]\big)$
Similarly,
$\vec{\text{CA}}.\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}-\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{a}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{c}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{c}}.\big(\vec{\text{c}}\times{\vec{\text{a}}}\big)$ (By distributive law)
$=\big[\vec{\text{a}}\vec{\text{a}}\vec{\text{b}}\big]+\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]-\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]-\big[\vec{\text{c}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{c}}\vec{\text{c}}\vec{\text{a}}\big]$
$=0\ \big(\therefore\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big)$
Hence, vector $\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is
perpendicular to all sides of $\triangle\text{ABC}$ and also perpendicular to the plane of $\triangle\text{ABC}.$

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Question 75 Marks

If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.

Answer

Let $\vec{\text{OA}}=4\vec{\text{i}}+3\vec{\text{j}}+3\vec{\text{k}},\vec{\text{OB}}=5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},\vec{\text{OC}}=5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}.$
$\therefore\vec{\text{AB}}=(5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}+(\text{x}-3)\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{AC}}=(5\hat{\text{i}}+3\hat{\text{j}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}-3\hat{\text{k}}$
$\vec{\text{AD}}=(7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=3\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Since the given four points are coplanar, so the vectors $\vec{\text{AB}},\vec{\text{AC}}$ and $\vec{\text{AD}}$ are also coplanar.
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=0$
$\begin{vmatrix}1&\text{x}-3&4\\1&0&-3\\3&3&-2 \end{vmatrix}=0$
$\Rightarrow 1(0+9)-(\text{x}-3)(-2+9)+4(3-0)=0$
$\Rightarrow 9-7\text{x}+21+12=0$
$\Rightarrow7\text{x}=42$
$\Rightarrow \text{x}=6$
Thus, the value of x is 6.

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Question 85 Marks

Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}},\vec{\text{b}}=5\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}$

Answer

Given:
$\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}$
$\vec{\text{b}}=5\hat{\text{k}}$
$\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}1&3&0\\0&0&5\\\lambda&-1&0 \end{vmatrix}=0$
$\Rightarrow1(0+5)-3(0-5\lambda)+0(0-0)=0$
$\Rightarrow5+15\lambda=0$
$\Rightarrow \lambda=-\frac{1}{3}$

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Question 95 Marks

Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}$

Answer

We know that three vectors $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&-2&3\\-2&3&-4\\1&-3&5 \end{vmatrix}$
$=1(15-12)+2(-10+4)+3(6-3)$
$=3-12+9$
$=0$
Hence, the given vectors are coplanar.

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Question 105 Marks

Show that the points A(-1, 4, -3), B(3, 2, -5), C(-3, 8, -5) and D(-3, 2, 1) are coplanar.

Answer

The points A, B, C and D will be coplanar off any one of the following triads of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}},$ etc.
To show that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplaner, we have to prove that their scaler triple product,
i.e., $\Big[\vec{\text{AB}}\vec{\text{ AC}}\vec{\text{ AD}}\Big]=0$
Now,
$\vec{\text{AB}}=\big[3-(-1)\big]\hat{\text{i}}+(2-4)\hat{\text{j}}+[-5-(-3)]\hat{\text{k}}\\=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{AC}}=[-3-1)]\hat{\text{i}}+(8-4)\hat{\text{j}}+[-5-(3)]\hat{\text{k}}\\=-2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{AD}}=[-3-(-1)]\hat{\text{i}}+(2-4)\hat{\text{j}}+[1-(3)]\hat{\text{k}}\\=-2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\Big[\vec{\text{AC}}\vec{\text{ AC}}\vec{\text{ AD}}\Big]=\begin{vmatrix}4&-2&-2\\-2&4&-2\\-2&-2&4 \end{vmatrix}$
$=4(16-4)+2(-8-4)-2(4+8)=0$
Thus, the given points are coplanar.

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