Vector - Collinear and Coplanar Vectors (Lecture-03)
Linear Combination of vectors
A vector $\overrightarrow{\mathrm{r}}$ is said to be a linear combination of vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ ………..etc. If there exist scalars $x$,
$y, z$ etc. such that $\quad \vec{r}=x \vec{a}+y \vec{b}+z \vec{c}+$ ………..
Collinearity and Coplanarity
(i) Two vectors $\vec{a} \& \vec{b}$ are collinear if and only if $\vec{a}=\lambda \vec{b}$ where $\lambda$ is a scalar
i.e.if $\vec{a}=a _{1} \hat{i}+a _{2} \hat{j}+a _{3} \hat{k}$ and $\vec{b}=b _{1} \hat{i}+b _{2} \hat{j}+b _{3} \hat{k}$ and $\vec{a} \& \vec{b}$ are collinear if $\dfrac{\mathrm{a} _{1}}{\mathrm{~b} _{1}}=\dfrac{\mathrm{a} _{2}}{\mathrm{~b} _{2}}=\dfrac{\mathrm{a} _{3}}{\mathrm{~b} _{3}}$.
Also if $\vec{a}$ and $\vec{b}$ are two non collinear vectors and $x \vec{a}+y \vec{b}=\overrightarrow{0}$, then $x=0$ and $y=0$
(ii) Three points with position vectors $\vec{a}, \vec{b}, \vec{c}$ are collinear if and only if there exist scalars $x, y, z$ not all zero such that $x \vec{a}+y \vec{b}+z \vec{c}=\overrightarrow{0}$ where $x+y+z=0$
Also three points $\mathrm{A}(\overrightarrow{\mathrm{a}}), \overrightarrow{\mathrm{B}}(\overrightarrow{\mathrm{b}}), \mathrm{C}(\overrightarrow{\mathrm{c}})$ are collinear if and only if $\overrightarrow{\mathrm{AB}}=\lambda \overrightarrow{\mathrm{AC}}$ for some scalar $\lambda$
(iii) Vectors lie on same or parallel planes are called coplanar vectors.
Let $\vec{a} \& \vec{b}$ be two given non-zero non-collinear vectors. Then any vector $\overrightarrow{\mathrm{r}}$ coplanar with $\overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{b}}$ can be uniquely expressed as $\overrightarrow{\mathrm{r}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$ for same scalars $\mathrm{x} \& \mathrm{y}$.
(iv) If three vectors are coplanar, then one of them can be expressed as a linear combination of the other two.
i.e. If $\vec{a}, \vec{b}, \vec{c}$ are coplanar then there exists scalars $x, y, z$ not all zero such that $x \vec{a}+y \vec{b}+z \vec{c}=\overrightarrow{0}$.
Also if $\vec{a}, \vec{b}, \vec{c}$ are non coplanar and $x \vec{a}+y \vec{b}+z \vec{c}=\overrightarrow{0}$, then $x=y=z=0$
(v) Four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar if and only if there exist scalars $x, y, z, u$ not all zero such that $x \vec{a}+y \vec{b}+z \vec{c}+u \vec{d}=\overrightarrow{0}$ where $x+y+z+u=0$
Note: If three non coplanar vectors $\vec{a}, \vec{b} \& \vec{c}$ are given, then every vector $\overrightarrow{\mathrm{r}}$ in space can be uniquely expressed as $\overrightarrow{\mathrm{r}}=x \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}+\mathrm{z} \overrightarrow{\mathrm{c}}$ for some scalars $x, y \& z$.
Linear independence and dependence of vectors
(i) A set of vectors $\vec{a} _{1}, \vec{a} _{2}, \ldots \ldots \ldots \ldots \ldots . . . \vec{a} _{n}$ is said to be linearly independent if $\mathrm{x} _{1} \overrightarrow{\mathrm{a}} _{1}+\mathrm{x} _{2} \overrightarrow{\mathrm{a}} _{2}+\ldots \ldots \ldots \ldots \ldots+\mathrm{x} _{\mathrm{n}} \overrightarrow{\mathrm{a}} _{\mathrm{n}}=\overrightarrow{0} \Rightarrow \mathrm{x} _{1}=\mathrm{x} _{2}=\ldots \ldots \ldots \ldots . .=\mathrm{x} _{\mathrm{n}}=0$
(ii) A set of vectors $\overrightarrow{\mathrm{a}} _{1}, \overrightarrow{\mathrm{a}} _{2}, \ldots \ldots \ldots \ldots \ldots . . \overrightarrow{\mathrm{a}} _{\mathrm{n}}$ is said to be linearly dependent if there exist scalars $\mathrm{x} _{1}, \mathrm{x} _{2} \ldots \ldots \ldots \mathrm{x} _{\mathrm{n}}$ not all zero such that $\mathrm{x} _{1} \overrightarrow{\mathrm{a}} _{1}+\mathrm{x} _{2} \overrightarrow{\mathrm{a}} _{2}+\ldots \ldots \ldots \ldots \ldots+\mathrm{x} _{\mathrm{n}} \overrightarrow{\mathrm{a}} _{\mathrm{n}}=\overrightarrow{0}$
i.e. a set of vectors is linearly dependent if and only if one of them is expressible as a linear
combination of others.
Note
(i) Two non-zero, non colinear vectors are linearly independent.
(ii) Any two collinear vectors are linearly dependent.
(iii) Any three non-coplanar vectors are linearly independent.
(iv) Any three coplanar vectors are linearly dependent.
(v) Any four vectors in 3-dimensional space are linearly dependent.
Solved Examples
1. If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors, no two of which are collinear, $\vec{a}+2 \vec{b}$ is collinear with $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}$ is collinear with $\overrightarrow{\mathrm{a}}$, then $|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}}|$ is equal to
(a) 0
(b) 1
(c) 9
(d) None of these
Show Answer
Solution
$\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{c}} \& \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}=\mu \overrightarrow{\mathrm{a}}$
$\Rightarrow \quad \overrightarrow{\mathrm{a}}-6 \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{c}}-2 \mu \overrightarrow{\mathrm{a}}$
$\Rightarrow \quad(1+2 \mu) \overrightarrow{\mathrm{a}}=(6+\lambda) \overrightarrow{\mathrm{c}}$
$\therefore \quad 1+2 \mu=0 \text { and } \lambda+6=0$ $\hspace {3cm}(\therefore \overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{c}}$ are non-collinear)
$\mu=-\dfrac{1}{2} \text { and } \lambda=-6$
$\Rightarrow \quad|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}}|=|-6 \overrightarrow{\mathrm{c}}+6 \overrightarrow{\mathrm{c}}|=0$
Answer (a)
2. If $\vec{x}$ and $\vec{y}$ are two non-collinear vectors and $A B C$ is a triangle with side lengths a,b,c satisfying $(20 a-15 b) \vec{x}+(15 b-12 c) \vec{y}+(12 c-20 a)(\vec{x} \times \vec{y})=\overrightarrow{0}$, then the triangle $A B C$ is
(a) acute angled
(b) obtuse angled
(c) right angled
(d) isosceles
Show Answer
Solution
$\therefore \overrightarrow{\mathrm{x}}, \overrightarrow{\mathrm{y}} \& \overrightarrow{\mathrm{x}} \times \overrightarrow{\mathrm{y}}$ are linearly independent $\Rightarrow 20 \mathrm{a}-15 \mathrm{~b}=15 \mathrm{~b}-12 \mathrm{c}=12 \mathrm{c}-20 \mathrm{a}=0$
$\Rightarrow \dfrac{\mathrm{a}}{3}=\dfrac{\mathrm{b}}{4}=\dfrac{\mathrm{c}}{5}=\lambda$
$\mathrm{a}=3 \lambda \mathrm{b}=4 \lambda \mathrm{c}=5 \lambda$
$\therefore$ the triangle is right angled.
Answer (c)
3. If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=4 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are linearly dependent vectors, and $|\overrightarrow{\mathbf{c}}|=\sqrt{3}$, then
(a) $\alpha=1, \beta=-1$
(b) $\alpha=1, \beta= \pm 1$
(c) $\alpha=-1, \beta= \pm 1$
(d) $\alpha= \pm 1, \beta=1$
Show Answer
Solution
$\because \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are linearly dependent
$\therefore\left[\begin{array}{lll}\overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{b}} & \overrightarrow{\mathrm{c}}\end{array}\right]=0$
$\Rightarrow\left|\begin{array}{lll}1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta\end{array}\right|=0$ gives $-\beta+1=0 \quad \Rightarrow \beta=1$
Also $|\overrightarrow{\mathrm{c}}|=\sqrt{3} \Rightarrow \sqrt{1+\alpha^{2}+\beta^{2}}=\sqrt{3}$
$\therefore \alpha= \pm 1$
Answer (d)
4. Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \mathrm{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}| ;|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$, and the angle between $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ is $30^{\circ}$; then $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})| \times \overrightarrow{\mathrm{c}}=$
(a) $2 / 3$
(b) $3 / 2$
(c) 2
(d) 3
Show Answer
Solution
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$ and $|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$ and $|\overrightarrow{\mathrm{c}}|^{2}+|\overrightarrow{\mathrm{a}}|^{2}-2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}})=8$
$\Rightarrow|\overrightarrow{\mathbf{c}}|^{2}+9-2|\overrightarrow{\mathbf{c}}|^{2}=8$
$\Rightarrow(|\overrightarrow{\mathbf{c}}|-1)^{2}=0$
$\Rightarrow|\overrightarrow{\mathrm{c}}|=1$
$\therefore|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=\sin 30^{\circ}$
$=3.1 . \dfrac{1}{2}=\dfrac{3}{2} \quad \quad \quad (\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=3)$
Answer (b)
5. Let $\vec{a}, \vec{b}, \vec{c}$ be vectors of equal magnitude such that angle between $\vec{a}$ and $\vec{b}$ is $\alpha, \vec{b}$ and $\vec{c}$ is $\beta, \vec{b}$ and $\vec{c}$ is $\gamma$. Then the minimum value of $\cos \alpha+\cos \beta+\cos \gamma$ is
(a) $\dfrac{1}{2}$
(b) $-\dfrac{1}{2}$
(c) $\dfrac{3}{2}$
(d) $-\dfrac{3}{2}$
Show Answer
Solution
Let $|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda$
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \alpha=\lambda^{2} \cos \alpha$
$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \beta=\lambda^{2} \cos \beta$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=|\overrightarrow{\mathrm{c}}||\overrightarrow{\mathrm{a}}| \cos \gamma=\lambda^{2} \cos \gamma$
Now $|\vec{a}+\vec{b}+\vec{c}|^{2} \geq 0$
$=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b}+\vec{c}+\vec{c} \cdot \vec{a}) \geq 0$
$\Rightarrow 3 \lambda^{2}+2 \lambda^{2}(\cos \alpha+\cos \beta+\cos \gamma) \geq 0$
$\Rightarrow \cos \alpha+\cos \beta+\cos \gamma \geq-\dfrac{3}{2}$
Answer (d)
6. If $\vec{a}=\hat{i}+\hat{j}, \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{r} \times \vec{a}=\vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b}=\vec{a} \times \vec{b}$ then a unit vector in the direction of $\overrightarrow{\mathrm{r}}$ is
(a) $\dfrac{1}{\sqrt{11}}(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$
(b) $\dfrac{1}{\sqrt{11}}(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
(c) $\dfrac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
(d) None of these
Show Answer
Solution
$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}=-\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=\overrightarrow{0}$
$\Rightarrow \overrightarrow{\mathrm{r}}$ is parallel to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$
$\therefore \overrightarrow{\mathrm{r}}=\lambda(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
also $|\overrightarrow{\mathrm{r}}|=\sqrt{11} \lambda$
$\therefore \hat{\mathrm{r}}=\dfrac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}=\dfrac{1}{\sqrt{11}}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
Answer (d)
Exercise
1.* Let $\vec{\alpha}=a \hat{i}+b \hat{j}+c \hat{k}, \vec{\beta}=b \hat{i}+c \hat{j}+a \hat{k}$ and $\vec{\gamma}=c \hat{i}+a \hat{j}+b \hat{k}$ be three coplanar vectors with $\mathrm{a} \neq \mathrm{b}$ and $\overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$. Then $\overrightarrow{\mathrm{v}}$ is perpendicular to
(a) $\vec{\alpha}$
(b) $\vec{\beta}$
(c) $\vec{\gamma}$
(d) None of these
Show Answer
Answer: a, b, c2. If $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors and $\gamma$ is a real number then the vectors $\vec{a}+2 \vec{b}+3 \vec{c}, \lambda \vec{b}+4 \vec{c}$ and $(2 \lambda-1) \overrightarrow{\mathrm{c}}$ are non coplanar for
(a) no value of $\lambda$
(b) all except one value of $\lambda$
(c) all except two values of $\lambda$
(d) all values of $\lambda$
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Answer: c3. $\vec{a}, \vec{b}, \vec{c}$ are three non coplanar vectors and $\vec{r}$ is any arbitrary vector, then $[\vec{b} \vec{c} \vec{r}] \vec{a}+[\vec{c} \vec{a} \vec{r}]$
$\overrightarrow{\mathrm{b}}+\left[\begin{array}{ll}\overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{b}} \ \mathrm{r}\end{array}\right] \overrightarrow{\mathrm{c}}$ is always equal to
(a) $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \vec{r}$
(b) $2[\vec{a} \quad \vec{b} \quad \vec{c}] \vec{r}$
(c) $3\left[\begin{array}{lll}\overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{b}} & \overrightarrow{\mathrm{c}}\end{array}\right] \overrightarrow{\mathrm{r}}$
(d) None of these
Show Answer
Answer: a4. The edges of a parallelopiped are of unit length and are parallel to non coplanar unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $\hat{a} \cdot \hat{b}=\hat{b} \cdot \hat{c}=\hat{c} \cdot \hat{a}=\dfrac{1}{2}$. Then the volume of the parallelopiped is
(a) $\dfrac{1}{\sqrt{2}}$ cu.unit
(b) $\dfrac{1}{2 \sqrt{2}}$ cu.unit
(c) $\dfrac{\sqrt{3}}{2}$ cu.unit
(d) $\dfrac{1}{\sqrt{3}}$ cu.unit
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Answer: a5. Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Then the vector $\overrightarrow{\mathrm{b}}$ satisfying $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3$ is
(a) $-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
(b) $2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
(c) $\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
(d) $\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
Show Answer
Answer: a6. $\mathrm{ABCD}$ is parallelogram. $\mathrm{A} _{1}$ and $\mathrm{B} _{1}$ are the mid points of sides $\mathrm{BC}$ and $\mathrm{CD}$ respectively. If $\overrightarrow{\mathrm{AA}} _{1}$ $+\overrightarrow{\mathrm{AB}} _{1}=\lambda \overrightarrow{\mathrm{AC}}$, then $\lambda$ is
(a) 1
(b) $\dfrac{3}{2}$
(c) $\dfrac{2}{3}$
(d) None of these
Show Answer
Answer: b7. $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=4 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{c}=\hat{i}+x _{1} \hat{j}+x _{2} \hat{k}$ are linearly dependent vectors and $|\vec{c}|=$ 2 , then the value of $729 \mathrm{x} _{1}{ }^{4}+343 \mathrm{x} _{1}{ }^{2} \mathrm{x} _{2}{ }^{2}+81 \mathrm{x} _{2}{ }^{2}+15=$
(a) 3689
(b) 3869
(c) 3698
(d) None of these
Show Answer
Answer: c8. $\vec{b}$ and $\vec{c}$ are non-collinear vectors. If $\vec{a} \times(\vec{b} \times \vec{c})+(\vec{b} \cdot \vec{a}) \vec{b}=(4-2 x-\operatorname{siny}) \vec{b}+\left(x^{2}-1\right) \vec{c}$ and $(\vec{c} \cdot \vec{c}) \vec{a}=\vec{c}$, then the value of $x^{\text {siny }}+(4 \sin y)^{4 x}+15$ must be
(a) 722
(b) 272
(d) 227
(d) None of these
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Answer: b9. If $\mathrm{G} _{1}, \mathrm{G} _{2}, \mathrm{G} _{3}$ are the centroids of the triangular faces $\mathrm{OBC}, \mathrm{OCA}, \mathrm{OAB}$ of a tetrahedron $\mathrm{OABC}$. If $\lambda$ be the ratio of the volume of the tetrahedron to the volume of the parallelepiped with $\mathrm{OG} _{1}, \mathrm{OG} _{2}, \mathrm{OG} _{3}$ as coterminus edges, the $\lambda=$
(a) $\dfrac{9}{4}$
(b) $\dfrac{4}{9}$
(c) $\dfrac{1}{6}$
(d) None of these
Show Answer
Answer: a10. Match the following:
| Column I | Column II | ||
|---|---|---|---|
| (a) | If $\vec{a}, \vec{b}$ and $\vec{c}$ are three mutually perpendicular vectors $\text { where }|\vec{a}|=|\vec{b}|=2,|\vec{c}|=1 \text {, then }\left[\begin{array}{lll} \vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a} \end{array}\right] \text { is }$ | (p) | -12 |
| (b) | If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at $\dfrac{\pi}{3}$, then $16\left[\begin{array}{lll}\vec{a} & \vec{b}+\vec{a} \times \vec{b} & \vec{b}\end{array}\right]$ is | (q) | 0 |
| (c) | If $\vec{b}$ and $\vec{c}$ are orthogonal unit vectors and $\vec{b} \times \vec{c}=\vec{a}$, then $\left[\begin{array}{lll}\vec{a}+\vec{b}+\vec{c} & \vec{a}+\vec{b} & \vec{b}+\vec{c}\end{array}\right]$ is | (r) | 16 |
| (d) | If $\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{a}\end{array}\right]=\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{b}\end{array}\right]=\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=0$ each vector being a non-zero vector, then $[\overrightarrow{\mathrm{x}} \overrightarrow{\mathrm{y}} \overrightarrow{\mathrm{c}}]$ is | (s) | 1 |
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Answer: $\mathrm{a} \rightarrow \mathrm{r} ; \mathrm{b} \rightarrow \mathrm{p} ; \dot{\mathrm{c}} \rightarrow \mathrm{s} ; \mathrm{d} \rightarrow \mathrm{q}$11. Read the passage and answer the following questions
Vectors $\vec{x}, \vec{y}$ and $\vec{z}$ each of the magnitude $\sqrt{2}$ make angle of $60^{\circ}$ with each other $\vec{x} \times(\vec{y} \times \vec{z})=\vec{a}, \vec{y} \times(\vec{z} \times \vec{x})=\vec{b}$ and $\vec{x} \times \vec{y}=\vec{c}$
(i) Vector $\overrightarrow{\mathrm{x}}$ is
(a) $\dfrac{1}{2}[(\vec{a}-\vec{b}) \times \vec{c}+(\vec{a}+\vec{b})]$
(b) $\dfrac{1}{2}[(\vec{a}+\vec{b}) \times \vec{c}+(\vec{a}-\vec{b})]$
(c) $\dfrac{1}{2}[-(\vec{a}+\vec{b}) \times \vec{c}+\vec{a}+\vec{b}]$
(d) $\dfrac{1}{2}[(\vec{a}+\vec{b}) \times \vec{c}-(\vec{a}+\vec{b})]$
(ii) Vector $\vec{y}$ is
(a) $\dfrac{1}{2}[(\vec{a}+\vec{c}) \times \vec{b}-\vec{b}-\vec{a}]$
(b) $\dfrac{1}{2}[(\vec{a}-\vec{c}) \times \vec{c}+\vec{b}+\vec{a}]$
(c) $\dfrac{1}{2}[(\vec{a}+\vec{b}) \times \vec{c}+\vec{b}+\vec{a}]$
(d) $\dfrac{1}{2}[(\vec{a}-\vec{c}) \times \vec{a}+\vec{b}-\vec{a}]$
(iii) Vector $\overrightarrow{\mathrm{z}}$ is
(a) $\dfrac{1}{2}[(\vec{a}-\vec{c}) \times \vec{c}-\vec{b}+\vec{a}]$
(b) $\dfrac{1}{2}[(\vec{b}+\vec{a}) \times \vec{c}+\vec{b}-\vec{a}]$
(c) $\dfrac{1}{2}[\vec{c} \times(\vec{a}-\vec{b})+\vec{b}+\vec{a}]$
(d) None of these.
BETA
