Vector - Vector Triple Product (Lecture-01)
Vector Triple Product
Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors, then the expression $\vec{a} \times(\vec{b} \times \vec{c})$ is a vector and is called a vector triple product.
Geometrically: $\vec{a} \times(\vec{b} \times \vec{c})$ is a vector that lies in the plane of $\vec{b}$ and $\vec{c}$ and perpendicular to $\vec{a}$.
- $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
- $(\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} . \vec{c}) \vec{b}-(\vec{b} . \vec{c}) \vec{a}$
- It is clear that $\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c}$
- $(\vec{a} \times \vec{b}) \times \vec{c}=\vec{a} \times(\vec{b} \times \vec{c})$ if and only if $\vec{b} \times(\vec{c} \times \vec{a})=\overrightarrow{0}$
- $(\vec{a} \times \vec{b}) \times \vec{a}=\vec{a} \times(\vec{b} \times \vec{a})=\vec{a} \times \vec{b} \times \vec{a}$
- $\hat{\mathrm{i}} \times(\hat{\mathrm{j}} \times \hat{\mathrm{k}})+\hat{\mathrm{j}} \times(\hat{\mathrm{k}} \times \hat{\mathrm{i}})+\hat{\mathrm{k}} \times(\hat{\mathrm{i}} \times \hat{\mathrm{j}})=\overrightarrow{0}$
- $\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})=2 \vec{a}$ where $\vec{a}$ is any vector.
Scalar Product of four vectors
$(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=\left|\begin{array}{ll}\vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d}\end{array}\right|$
Vector product of four Vectors
$(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\left\{\begin{array}{l}{[\vec{a} \vec{b} \vec{d}] \vec{c}-[\vec{a} \vec{b} \vec{c}] \vec{d}} \\ {[\vec{a} \vec{c} \vec{d}] \vec{b}-[\vec{b} \vec{c} \vec{d}] \vec{a}}\end{array}\right\}$
-
$\left[\begin{array}{lll} \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} & \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \end{array}\right]=\left[\begin{array}{lll} \overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{b}} & \overrightarrow{\mathrm{c}} \end{array}\right]^2$
-
$\quad(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}$ if $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar.
Reciprocal system of Vectors
If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{a}^{\prime}, \vec{b}^{\prime}, \vec{c}^{\prime}$ are two sets of non coplanar vectors such that $\vec{a} \cdot \vec{a}^{\prime}=\vec{b} \cdot \vec{b}^{\prime}=\vec{c} \cdot \vec{c}^{\prime}$ $=1$, then the two systems are called reciprocal system of vectors.
Note :
$\vec{a}^{\prime}=\dfrac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \quad \vec{b}^{\prime}=\dfrac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \quad \vec{c}=\dfrac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$
Properties
(i) $\overrightarrow{\mathrm{a}}=\dfrac{\overrightarrow{\mathrm{b}^{\prime}} \times \overrightarrow{\mathrm{c}}^{\prime}}{\left[\overrightarrow{\mathrm{a}}^{\prime} \overrightarrow{\mathrm{b}}^{\prime} \overrightarrow{\mathrm{c}}^{\prime}\right]} ; \overrightarrow{\mathrm{b}}=\dfrac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}^{\prime}}{\left[\overrightarrow{\mathrm{a}}^{\prime} \overrightarrow{\mathrm{b}}^{\prime} \mathrm{c}^{\prime}\right]} ; \overrightarrow{\mathrm{c}}=\dfrac{\overrightarrow{\mathrm{a}^{\prime} \times \overrightarrow{\mathrm{b}}^{\prime}}}{\left[\overrightarrow{\mathrm{a}}^{\prime} \overrightarrow{\mathrm{b}}^{\prime} \mathrm{c}^{\prime}\right]}$
i.e. reciprocal vectors of $\vec{a}^{\prime}, \vec{b}^{\prime}, \vec{c}^{\prime}$ are $\vec{a}, \vec{b}, \vec{c}$
(ii) $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]\left[\begin{array}{lll}\vec{a}^{\prime} & \vec{b}^{\prime} & \vec{c}^{\prime}\end{array}\right]=1$
(iii) $\vec{a} \cdot \vec{b}^{\prime}=\vec{a} \cdot \vec{c}^{\prime}=\vec{b} \cdot \vec{a}^{\prime}=\vec{b} \cdot \vec{c}^{\prime}=\vec{c} \cdot \vec{a}^{\prime}=\vec{c} \cdot \vec{b}^{\prime}=0$
(iv) $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}^{\prime}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}^{\prime}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}^{\prime}=\overrightarrow{0}$
(v) Orthonormal Vectors $\hat{i}, \hat{j}, \hat{k}$ are self reciprocal vectors.
Solved Examples
1. The value of $\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})$ is
(a) $\vec{a}+\vec{b}+\vec{c}$
(b) $3(\vec{b} \cdot \vec{c}) \vec{a}$
(c) $\overrightarrow{0}$
(d) $\vec{a} \cdot \vec{b}+\vec{b} \vec{c}+\vec{c} \cdot \vec{a}$
Show Answer
Solution
$\begin{aligned} & \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} \\ & \vec{b} \times(\vec{c} \times \vec{a})=(\vec{b} \cdot \vec{a}) \vec{c}-(\vec{b} \cdot \vec{c}) \vec{a} \\ & \vec{c} \times(\vec{a} \times \vec{b})=(\vec{c} \cdot \vec{b}) \vec{a}-(\vec{c} \cdot \vec{a}) \vec{b} \end{aligned}$
Adding all three equations, we get
$\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a}) \vec{c} \times(\vec{a} \times \vec{b})=\overrightarrow{0}$
Answer (c)
2. $\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})$ is equal to
(a) $3 \vec{a}$
(b) $2 \vec{a}$
(c) $\overrightarrow{0}$
(d) $\vec{a}$
Show Answer
Solution
$\begin{aligned} & \hat{i} \times(\vec{a} \times \hat{i})=(\hat{i} \cdot \hat{i}) \vec{a}-(\hat{i} \cdot \vec{a}) \hat{i} \\ & \hat{j} \times(\vec{a} \times \hat{j})=(\hat{j} \cdot \hat{j}) \vec{a}-(\hat{j} \cdot \vec{a}) \hat{j} \\ & \hat{k} \times(\vec{a} \times \hat{k})=(\hat{k} \cdot \hat{k}) \vec{a}-(\hat{k} \cdot \vec{a}) \hat{k} \end{aligned}$
On Adding, we get
$$ \begin{aligned} \hat{\mathrm{i}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}})+\hat{\mathrm{j}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}})+\hat{\mathrm{k}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}) & =3 \overrightarrow{\mathrm{a}}-{(\hat{\mathrm{i}} \cdot \overrightarrow{\mathrm{a}}) \hat{\mathrm{i}}+(\hat{\mathrm{j}} \cdot \overrightarrow{\mathrm{a}}) \hat{\mathrm{j}}+(\hat{\mathrm{k}} \cdot \overrightarrow{\mathrm{a}}) \hat{\mathrm{k}}} \\ & =3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \\ & =2 \vec{a} \end{aligned} $$
Answer (b)
3. $\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right]=$
(a) $3\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$
(b) $2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$
(c) $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$
(d) $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^{2}$
Show Answer
Solution
$\begin{aligned} & (\vec{a} \times \vec{b}) .\{(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\} \\ & =(\vec{a} \times \vec{b}) .\left\{\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right] \vec{c}-0\right\} \\ & =\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{a} \end{array}\right]\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right] \\ & =\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right] \end{aligned}$
$=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]^{2}$
Answer (d)
4. If $\vec{a}=3 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=-\hat{i}+\hat{j}+3 \hat{k}$ then which of the following in true.
(a) $(\vec{a} \times \vec{b}) \cdot \vec{c}=30$
(b) $(\vec{a} \cdot \vec{b}) \times \vec{c}=\hat{i}-\hat{j}-3 \hat{k}$
(c) $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
(d) $(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{a}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Show Answer
Solution
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ 1 & 3 & -1 \end{array}\right|=-2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$
$(\vec{a} \times \vec{b}) \cdot \vec{c}=2+4+30=36$
(a) is wrong.
(b) is meaningless.
$\begin{aligned} & \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} \\ & =-\overrightarrow{\mathrm{b}}-(-1) \overrightarrow{\mathrm{c}} \\ & =-\vec{b}+\vec{c} \\ & =-2 \hat{i}-2 \hat{j}+4 \hat{k} \\ & \text { (c) is wrong } \end{aligned}$
$(\vec{b} \times \vec{c}) \times \vec{a}=-{\vec{a} \times(\vec{b} \times \vec{c})}$
$=-(-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Answer (d)
5. If $\vec{a}, \vec{b}$, $\vec{c}$ be three non parallel unit vectors such that $\vec{a} \times(\vec{b} \times \vec{c})=\dfrac{1}{2} \vec{b}$, then angle between $\vec{a}$ and $\vec{c}$ is
(a) $\dfrac{\pi}{2}$
(b) 0
(c) $\dfrac{\pi}{6}$
(d) $\dfrac{\pi}{3}$
Show Answer
Solution
$\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\dfrac{1}{2} \vec{b}-0 \vec{c}$
$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\dfrac{1}{2}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$
$|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}| \cos \theta=\dfrac{1}{2}$
(where $\theta$ is the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}}$ )
$\begin{aligned} \cos \theta & =\dfrac{1}{2} \\ \theta & =\dfrac{\pi}{3} \end{aligned}$
Answer (d)
6. $\vec{a} \times(\vec{a} \times(\vec{a} \times \vec{b}))$ is equal to
(a) $(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{a})$
(b) $(\vec{a} \cdot \vec{b})(\vec{b} \times \vec{a})$
(c) $(\vec{b} \cdot \vec{b})(\vec{b} \times \vec{a})$
(d) None of these
Show Answer
Solution
$\begin{aligned} \vec{a} \times(\vec{a} \times(\vec{a} \times \vec{b})) & =\vec{a} \times((\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b} \\ & =(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{a})-(\vec{a} \cdot \vec{a})(\vec{a} \times \vec{b}) \\ & =\overrightarrow{0}+(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{a}) \\ & =(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{a}) \end{aligned}$
Answer (a)
7. Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. 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 $30^{\circ}$ the $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|=$
(a) $\dfrac{2}{3}$
(b) $\dfrac{3}{2}$
(c) 2
(d) 3
Show Answer
Solution
$|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$
Squaring both side
$\begin{aligned} & |\overrightarrow{\mathrm{c}}|^{2}+|\overrightarrow{\mathrm{a}}|^{2}-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=8 \\ & |\overrightarrow{\mathrm{c}}|^{2}+9-2|\overrightarrow{\mathrm{c}}|=8 \\ & |\overrightarrow{\mathrm{c}}|^{2}-2|\overrightarrow{\mathrm{c}}|+1=0 \\ & (|\overrightarrow{\mathrm{c}}|-1)^{2}=0 \\ & |\overrightarrow{\mathrm{c}}|=1 \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$
$\therefore|\vec{a} \times \vec{b}|=3$
$|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}$
$=3.1 . \dfrac{1}{2}$
$=\dfrac{3}{2}$
Answer (b)
Exercise
1. If $\overrightarrow{\mathrm{u}} . \overrightarrow{\mathrm{v}}, \overrightarrow{\mathrm{w}}$ are three non coplanar unit vectors and $\alpha, \beta, \gamma$ are the angles between $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}$; and $\vec{v}$ and $\vec{w} ; \vec{w}$ and $\vec{u}$ respectively and $\vec{x}, \vec{y}, \vec{z}$ are unit vectors along the bisectors of the angles $\alpha, \beta, \gamma$ respectively. If $\left[\begin{array}{lll}\vec{x} \times \vec{y} & \vec{y} \times \vec{z} & \vec{z} \times \vec{x}\end{array}\right]=k\left[\begin{array}{ll}\vec{u} \quad \vec{v} & \vec{w}\end{array}\right]^{2} \sec ^{2} \dfrac{\alpha}{2} \sec ^{2} \dfrac{\beta}{2} \sec ^{2} \dfrac{\gamma}{2}$ then $\mathrm{k}=$
(a) 2
(b) 16
(c) $\dfrac{1}{16}$
(d) None of these
Show Answer
Answer: c2. A mirror and a source of light are situated at the origin $\mathrm{O}$ and a point on $\mathrm{OX}$ respectively. A ray of light from the source strikes the mirror and is reflected. If d. $\mathrm{r}^{\prime} \mathrm{s}$ of the normal to the plane of mirror are $1,-1,1$, then the d. $\mathrm{c}^{\prime} \mathrm{s}$ of the reflected ray are
(a) $\dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3}$
(b) $\dfrac{1}{3}, \dfrac{-2}{3}, \dfrac{-2}{3}$
(c) $\dfrac{-1}{3}, \dfrac{-2}{3}, \dfrac{-2}{3}$
(d) $\dfrac{-1}{3}, \dfrac{-2}{3}, \dfrac{2}{3}$
Show Answer
Answer: d3. Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the points $A, B, C$ respectively. Let $\alpha, \beta$ and $\gamma$ the inclinations between $\vec{b}, \vec{c} ; \vec{a}, \vec{b}$ and $\vec{a}, \vec{c}$. Then the volume $V$ of the tetrahedron OABC is given by
(a) $\mathrm{V}^{2}=\dfrac{\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}}{36}\left|\begin{array}{ccc}1 & \cos \beta & \cos \gamma \\ \cos \beta & 1 & \cos \alpha \\ \cos \gamma & \cos \alpha & 1\end{array}\right|$
(b) $\mathrm{V}^{2}=\dfrac{\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}}{6}\left|\begin{array}{ccc}1 & \cos \beta & \cos \gamma \\ \cos \beta & 1 & \cos \alpha \\ \cos \gamma & \cos \alpha & 1\end{array}\right|$
(c) $\mathrm{V}^{2}=\dfrac{\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}}{36}\left|\begin{array}{ccc}0 & \cos \beta & \cos \gamma \\ \cos \beta & 0 & \cos \alpha \\ \cos \gamma & \cos \alpha & 0\end{array}\right|$
(d) $\mathrm{V}^{2}=\dfrac{\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}}{36}\left|\begin{array}{ccc}1 & \sin \beta & \sin \gamma \\ \sin \beta & 1 & \sin \alpha \\ \sin \gamma & \sin \alpha & 1\end{array}\right|$
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Answer: a4.* $\vec{a}$ and $\vec{b}$ are two given vectors. On these vectors as adjacent sides a parallelogram is constructed. The vector which is the altitude of the parallelogram and which is perpendicular to $\vec{a}$ is
(a) $\dfrac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^{2}} \vec{a}-\vec{b}$
(b) $\dfrac{1}{|\vec{a}|^{2}}\left\{|\vec{a}|^{2} \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a}\right\}$
(c) $\dfrac{(\vec{a} \cdot \vec{b})}{|\vec{b}|^{2}} \vec{a}-\vec{b}$
(d) None of these
Show Answer
Answer: a, b5. Let $\vec{a}$ and $\vec{b}$ be two non-collinear unit vectors. If $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b})$ and $\vec{v}=\vec{a} \times \vec{b}$, then $|\vec{v}|$ is
(a) $|\overrightarrow{\mathrm{u}}|$
(b) $|\overrightarrow{\mathrm{u}}|+|\overrightarrow{\mathrm{u}} . \overrightarrow{\mathrm{a}}|$
(c) $|\overrightarrow{\mathrm{u}}|+|\overrightarrow{\mathrm{u}} \cdot \overrightarrow{\mathrm{b}}|$
(d) $|\overrightarrow{\mathrm{u}}|+\overrightarrow{\mathrm{u}} .(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$
Show Answer
Answer: c6. Let $\vec{a}=2 \hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+4 \hat{k}$. A vector $\vec{r}$ satisfying $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{a}}=0$ is
(a) $-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
(b) $-2 \hat{i}+\hat{j}+3 \hat{k}$
(c) $-2 \hat{i}-\hat{j}+5 \hat{k}$
(d) $\hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Show Answer
Answer: d7. Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and let $\overrightarrow{\mathrm{r}}$ be a variable vector such that $\overrightarrow{\mathrm{r}} . \hat{\mathrm{i}}, \overrightarrow{\mathrm{r}} . \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{r}} . \hat{\mathrm{k}}$ are positive integers. If $\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{a}} \leq 12$, then the total number of such vectors is
(a) ${ }^{12} \mathrm{C} _{9}-1$
(b) ${ }^{12} \mathrm{C} _{3}$
(c) ${ }^{12} \mathrm{C} _{8}$
(d) None of these
Show Answer
Answer: b8. If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest value of $|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|$ is
(a) 2
(b) 4
(c) $2 \sqrt{2}$
(d) $\sqrt{2}$
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Answer: c9.* If the magnitude of the moment about the point $\hat{j}+\hat{k}$ of a force $\hat{i}+\alpha \hat{j}-\hat{k}$ acting through the point $\hat{i}+\hat{j}$ is $\sqrt{8}$, then the value of $\alpha$ is
(a) 1
(b) 2
(c) 3
(d) -2
Show Answer
Answer: b, d10. Read the paragraph and answer the following questions.
Let $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}, \vec{a} _{1}=2 \hat{i}-3 \hat{j}+6 \hat{k}$ and $\vec{a} _{2}=-2 \hat{i}+3 \hat{j}+6 \hat{k}$.
Let $\vec{a} _{1}$ be the projection of $\vec{a}$ and $\vec{b}$ the projection of $\vec{a} _{1}$ on $\vec{c}$, then
(i) $\overrightarrow{\mathrm{a}} _{2}$ is equal to
(a) $\dfrac{943}{49}(2 \hat{i}-3 \hat{j}-6 \hat{k})$
(b) $\dfrac{943}{49^{2}}(2 \hat{i}-3 \hat{j}-6 \hat{k})$
(c) $\dfrac{943}{49}(-2 \hat{i}+3 \hat{j}+6 \hat{k})$
(d) $\dfrac{943}{49^{2}}(-2 \hat{i}+3 \hat{j}+6 \hat{k})$
(ii) $\overrightarrow{\mathrm{a}} _{1} \cdot \overrightarrow{\mathrm{b}}$ is equal to
(a) -41
(b) $\dfrac{-41}{7}$
(c) 41
(d) 287
(iii) Which of the following is true?
(a) $\vec{a} \& \vec{a} _{2}$ are collinear
(b) $\overrightarrow{\mathrm{a}} _{1}$ and $\overrightarrow{\mathrm{c}}$ are collinear
(c) $\vec{a}, \vec{a} _{1}, \vec{b}$ are coplanar
(d) $\vec{a}, \vec{a} _{1}, \vec{a} _{2}$ are coplanar
Show Answer
Answer: (i) b (ii) a (iii) c11. Match the following.
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ unit vectors such that $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=\cos \theta, \vec{d} \cdot \vec{a}=\vec{d} \cdot \vec{b}=\vec{d} \cdot \vec{c}=\cos \alpha$. Then
Column I | Column II | ||
---|---|---|---|
(a) | $\dfrac{\pi}{2}$ | (p) | $\dfrac{\pi}{4}$ |
(b) | $\dfrac{\pi}{3}$ | (q) | $\cos ^{-1} \dfrac{1}{\sqrt{3}}$ |
(c) | $\dfrac{2 \pi}{3}$ | (r) | $\cos ^{-1} \sqrt{\dfrac{2}{3}}$ |
(d) | $\cos ^{-1} \dfrac{1}{4}$ | (s) | $\dfrac{\pi}{2}$ |
Show Answer
Answer: $\mathrm{a} \rightarrow \mathrm{q} ; \mathrm{b} \rightarrow \mathrm{r} ; \mathrm{c} \rightarrow \mathrm{s} ; \mathrm{d} \rightarrow \mathrm{p}$12. If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are the unit vectors such that $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$ and $\vec{a} \cdot \vec{c}=\dfrac{1}{2}$, then
(a) $\vec{a}, \vec{b}, \vec{c}$ are non coplanar
(b) $\vec{a}, \vec{b}, \vec{d}$ are non coplanar
(c) $\vec{b}, \vec{d}$ are non parallel
(d) $\vec{a}, \vec{d}$ are parallel and $\vec{b}, \vec{c}$ are parallel