Vector Algebra
Let $\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b} \cdot \vec{c}=0$. Consider the following two statements:
(A) $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ for all $\lambda \in \mathbb{R}$.
(B) $\vec{a}$ and $\vec{c}$ are always parallel.
Then,
If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$,
then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to :
If $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three non-zero vectors and $\widehat n$ is a unit vector perpendicular to $\overrightarrow c $ such that $\overrightarrow a = \alpha \overrightarrow b - \widehat n,(\alpha \ne 0)$ and $\overrightarrow b \,.\overrightarrow c = 12$, then $\left| {\overrightarrow c \times (\overrightarrow a \times \overrightarrow b )} \right|$ is equal to :
Let a unit vector $\widehat{O P}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the co-ordinate axes $\mathrm{OX}$, $\mathrm{OY}, \mathrm{OZ}$ respectively, where $\beta \in\left(0, \frac{\pi}{2}\right)$. If $\widehat{\mathrm{OP}}$ is perpendicular to the plane through points $(1,2,3),(2,3,4)$ and $(1,5,7)$, then which one of the following is true?
If $\overrightarrow a = \widehat i + 2\widehat k,\overrightarrow b = \widehat i + \widehat j + \widehat k,\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k,\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0 $ and $\overrightarrow r \,.\,\overrightarrow a = 0$. Then $\overrightarrow r \,.\,\overrightarrow c $ is equal to :
Let $\overrightarrow a = 4\widehat i + 3\widehat j$ and $\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k$. If $\overrightarrow c $ is a vector such that $\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right) + 25 = 0,\overrightarrow c \,.(\widehat i + \widehat j + \widehat k) = 4$, and projection of $\overrightarrow c $ on $\overrightarrow a $ is 1, then the projection of $\overrightarrow c $ on $\overrightarrow b $ equals :
If the vectors $\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k$, $\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k$ and $\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k$ are coplanar and the projection of $\overrightarrow a $ on the vector $\overrightarrow b $ is $\sqrt {54} $ units, then the sum of all possible values of $\lambda + \mu $ is equal to :
Let $\overrightarrow a = - \widehat i - \widehat j + \widehat k,\overrightarrow a \,.\,\overrightarrow b = 1$ and $\overrightarrow a \times \overrightarrow b = \widehat i - \widehat j$. Then $\overrightarrow a - 6\overrightarrow b $ is equal to :
If the four points, whose position vectors are $3\widehat i - 4\widehat j + 2\widehat k,\widehat i + 2\widehat j - \widehat k, - 2\widehat i - \widehat j + 3\widehat k$ and $5\widehat i - 2\alpha \widehat j + 4\widehat k$ are coplanar, then $\alpha$ is equal to :
The vector $\overrightarrow a = - \widehat i + 2\widehat j + \widehat k$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $\overrightarrow b $. Then the projection of $3\overrightarrow a + \sqrt 2 \overrightarrow b $ on $\overrightarrow c = 5\widehat i + 4\widehat j + 3\widehat k$ is :
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three non zero vectors such that $\overrightarrow b $ . $\overrightarrow c $ = 0 and $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = {{\overrightarrow b - \overrightarrow c } \over 2}$. If $\overrightarrow d $ be a vector such that $\overrightarrow b \,.\,\overrightarrow d = \overrightarrow a \,.\,\overrightarrow b $, then $(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d )$ is equal to
Let $\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k$ and $\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k$. Let ${\overrightarrow \beta _1}$ be parallel to $\overrightarrow \alpha $ and ${\overrightarrow \beta _2}$ be perpendicular to $\overrightarrow \alpha $. If $\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2}$, then the value of $5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right)$ is :
Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that
${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$. Then ${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$ is equal to :
Let $\overrightarrow u = \widehat i - \widehat j - 2\widehat k,\overrightarrow v = 2\widehat i + \widehat j - \widehat k,\overrightarrow v .\,\overrightarrow w = 2$ and $\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v $. Then $\overrightarrow u .\,\overrightarrow w $ is equal to :
Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^{2}=50$, then $|72-| \vec{b}+\left.\vec{c}\right|^{2} \mid$ is equal to __________.
Explanation:
Given that $\vec{a} = \vec{b} \times \vec{c}$, we can find the magnitudes of $\vec{a}$ and $\vec{c}$:
$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$
$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$
We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :
$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$
Plugging in the known values :
$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$
Solving for the sine of the angle between the vectors :
$\sin\theta = \frac{1}{10}$
Now we can find $|\vec{b} + \vec{c}|^2$ using the formula :
$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$
We have the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$, and we can use the relationship between sine and cosine: $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$.
Substitute the values into the formula :
$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$
Finally, we need to find the absolute value of the difference :
$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$
Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11, \vec{b} \cdot(\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$, then $|\vec{a} \times \vec{c}|^{2}$ is equal to _________.
Explanation:
$ \begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ & \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\ & \vec{a} \cdot \vec{c}=11 \\\\ & \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\ & \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\ & (\vec{b} \times \vec{a}) \cdot \vec{c}=27 \end{aligned} $
$ \text { Let } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} $
As $\vec{a} \cdot \vec{c}=11$
$ \therefore $ $ c_1+2 c_2+3 c_3=11 $ ......(i)
Also, $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$
$ \begin{aligned} & \therefore c_1+c_2-c_3=-\sqrt{3} \sqrt{3} \\\\ & \Rightarrow c_1+c_2-c_3=-3 ......(ii) \end{aligned} $
Also, $\vec{b} \cdot(\vec{a} \times \vec{c})=27$
$ \therefore $ $ 5 c_1-4 c_2+c_3=27 $ ...........(iii)
From (i), (ii) & (iii)
$ \vec{c}=3 \hat{i}-2 \hat{j}+4 \hat{k} $
$ \begin{aligned} & |\vec{a} \times \vec{c}|^2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & +4 \end{array}\right|^2 \\\\ & =|14 \hat{i}+5 \hat{j}-8 \hat{k}|^2 \\\\ & =14^2+5^2+8^2=285 \end{aligned} $
Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If
$\vec{a} \cdot \vec{c}=-12, \vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$, then $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to _______________.
Explanation:
Now, $\vec{a} \cdot \vec{c}=-12$
$ \Rightarrow 6 c_1+9 c_2+12 c_3=-12 $ ..............(i)
Also, $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$
$ \Rightarrow c_1-2 c_2+c_3=5 $ ................(ii)
$ \begin{aligned} & \text { Now, } \vec{a} \times \vec{c}=\vec{a} \times \vec{b} \\\\ & \Rightarrow \vec{a} \times(\vec{c}-\vec{b})=0 \\\\ & \Rightarrow \vec{a} \text { is parallel to }(\vec{c}-\vec{b}) \\\\ & \Rightarrow \vec{a}=\lambda(\vec{c}-\vec{b}) \\\\ & \Rightarrow 6 \hat{i}+9 \hat{j}+12 \hat{k}=\lambda\left(c_1-\alpha\right) \hat{i}+\lambda\left(c_2-11\right) \hat{j}+\lambda\left(c_3+2\right) \hat{k} \end{aligned} $
On comparing, we get
$ c_1=\frac{6}{\lambda}+\alpha, c_2=\frac{9}{\lambda}+11, c_3=\frac{12}{\lambda}-2 $
Put there values in (ii), we get
$ \begin{aligned} & \frac{6}{\lambda}+\alpha-\frac{18}{\lambda}-22+\frac{12}{\lambda}-2=5 \\\\ & \Rightarrow \alpha=29 \end{aligned} $
From (i) and values of $\mathrm{c}_1, \mathrm{c}_2, \mathrm{c}_3$, and $\alpha$ we have
$ \begin{aligned} & 6\left(\frac{6}{\lambda}+29\right)+9\left(\frac{9}{\lambda}+11\right)+12\left(\frac{12}{\lambda}-2\right)=-12 \\\\ & \Rightarrow \frac{261}{\lambda}=-261 \Rightarrow \lambda=-1 \end{aligned} $
$ \begin{aligned} & \text { So, } c_1=23, c_2=2, c_3=-14 \\\\ & \therefore \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=(23 \hat{i}+2 \hat{j}+-14 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\\\ & =23+2-14=11 \end{aligned} $
Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$ and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha>0$. If the minimum value of the scalar triple product $\left[ {\matrix{ {\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr } } \right]$ is $-\alpha \sqrt{3401}$, and $|\vec{u} \cdot \hat{i}|^{2}=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m+n$ is equal to ____________.
Explanation:
$\left[ {\matrix{ {\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr } } \right] = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$
$=|\vec{u}||\vec{v} \times \vec{w}| \times \cos \theta$
$=\alpha \sqrt{34 \alpha^{2}+1} \cos \theta$
$[\vec{u} \vec{v} \vec{w}]_{\min }=-\alpha \sqrt{3401}$
$\alpha \sqrt{34 \alpha^{2}+1} \times(-1)=-\alpha \sqrt{3401}$
(taking $\cos \theta=1$ )
$\Rightarrow \alpha=10$
$\vec{v} \times \vec{w}=\hat{i}-50 \hat{j}-30 \hat{k}$
$\cos \theta=-1 \Rightarrow \vec{u}$ is antiparallel to $\vec{v} \times \vec{w}$
$\vec{u}=-|\vec{u}| \cdot \frac{\vec{v} \times \vec{w}}{|\vec{v} \times \vec{w}|}=\frac{-10(\hat{i}-50 \hat{j}-30 \hat{k})}{\sqrt{3401}}$
$|\vec{u} \cdot \hat{i}|^{2}=\left|\frac{-10}{\sqrt{3401}}\right|^{2}=\frac{100}{3401}=\frac{m}{n}$
$m+n=3501$
$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$ and $D(4,5,0),|\lambda| \leq 5$ are the vertices of a quadrilateral $A B C D$. If its area is 18 square units, then $5-6 \lambda$ is equal to __________.
Explanation:
$ \begin{aligned} & =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) \\\\ & \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} \\\\ & =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 \\\\ & =\lambda^2+10 \lambda+9=0 \\\\ & =\lambda=-1,-9 \\\\ & |\lambda| \leq 5 \Rightarrow \lambda=-1 \\\\ & 5-6 \lambda=5-6(-1)=11 \end{aligned} $
$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$.
If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^{2}$ is equal to __________.
Explanation:
$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
$ \begin{aligned} & \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\ & |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\ & 31=31 \lambda^{2} \\\\ & \lambda=\pm 1 \\\\ & \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\ & \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\ & |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\ & =\frac{3}{4} \\\\ & \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\ & \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3 \end{aligned} $
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^{2}$ is equal to ___________.
Explanation:
$ \begin{aligned} & \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\ & \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\ & \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\ &=36 \end{aligned} $
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three non-zero non-coplanar vectors. Let the position vectors of four points $A,B,C$ and $D$ be $\overrightarrow a - \overrightarrow b + \overrightarrow c ,\lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c , - \overrightarrow a + 2\overrightarrow b - 3\overrightarrow c $ and $2\overrightarrow a - 4\overrightarrow b + 6\overrightarrow c $ respectively. If $\overrightarrow {AB} ,\overrightarrow {AC} $ and $\overrightarrow {AD} $ are coplanar, then $\lambda$ is equal to __________.
Explanation:
$ \overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c} $
$ \overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c} $
$ \left|\begin{array}{ccc} \lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|=0 $
$ \Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0 $
$ \Rightarrow(\lambda-1)=1 \Rightarrow \lambda=2 $
Let $\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow c $. Then $\left| {\overrightarrow a \,.\,\overrightarrow b } \right|$ is equal to :
Explanation:
Now $\vec{a} \cdot \overrightarrow{\mathbf{c}}=7$ gives $2 \lambda^2+12=7 \mu$
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$
$\mu=2$ and $\lambda^2=1$
$ |\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8 $
Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168$, then $|\vec{a}|+|\vec{b}|+|\vec{c}|$ is equal to :
Let $\overrightarrow{\mathrm{a}}=3 \hat{i}+\hat{j}$ and $\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}+\hat{k}$. Let $\overrightarrow{\mathrm{c}}$ be a vector satisfying $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$. If $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are non-parallel, then the value of $\lambda$ is :
Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $\theta$ is the angle between the vectors $(\hat{a}+\hat{b})$ and $(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$, then the value of $164 \,\cos ^{2} \theta$ is equal to :
Let S be the set of all a $\in R$ for which the angle between the vectors $ \vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$ and $\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$, $(b>1)$ is acute. Then S is equal to :
Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$ be such that for $\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$. Then, the set of all values of $t$ is :
Let a vector $\vec{a}$ has magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$, the vector $(x \vec{a}+y \vec{b})$ is perpendicular to the vector $(6 y \vec{a}-18 x \vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to :
Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors, such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to :
$ \text { Let } \vec{a}=2 \hat{i}-\hat{j}+5 \hat{k} \text { and } \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k} \text {. If }((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2} \text {, then }|\vec{b} \times 2 \hat{j}| $ is equal to :
Let $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0$. If the projection of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ on the vector $-\hat{i}+2 \hat{j}-2 \hat{k}$ is 30, then $\alpha$ is equal to :
Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and let $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :
Let $\mathrm{ABC}$ be a triangle such that $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$ and $\vec{b} \cdot \vec{c}=12$. Consider the statements :
$(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)$
$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
Then
Let a vector $\overrightarrow c $ be coplanar with the vectors $\overrightarrow a = - \widehat i + \widehat j + \widehat k$ and $\overrightarrow b = 2\widehat i + \widehat j - \widehat k$. If the vector $\overrightarrow c $ also satisfies the conditions $\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$ and $\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$, then the value of $|\overrightarrow c {|^2}$ is equal to :
$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$
$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$
$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$
If $\alpha$ is the smallest positive integer for which $\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $ are noncollinear, then the length of the median, in $\Delta$ABC, through A is :
Let $\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$, $\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$ and $\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$ where $\alpha ,\,\beta \in R$, be three vectors. If the projection of $\overrightarrow a $ on $\overrightarrow c $ is ${{10} \over 3}$ and $\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$, then the value of $\alpha + \beta $ is equal to :
Let $\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$, where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow a $ and $\overrightarrow b $ is $\sqrt {15({\alpha ^2} + 4)} $, then the value of $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a \,.\,\overrightarrow b } \right){\left| {\overrightarrow b } \right|^2}$ is equal to :
Let $\overrightarrow a $ be a vector which is perpendicular to the vector $3\widehat i + {1 \over 2}\widehat j + 2\widehat k$. If $\overrightarrow a \times \left( {2\widehat i + \widehat k} \right) = 2\widehat i - 13\widehat j - 4\widehat k$, then the projection of the vector $\overrightarrow a $ on the vector $2\widehat i + 2\widehat j + \widehat k$ is :
Let $\overrightarrow a $ and $\overrightarrow b $ be the vectors along the diagonals of a parallelogram having area $2\sqrt 2 $. Let the angle between $\overrightarrow a $ and $\overrightarrow b $ be acute, $|\overrightarrow a | = 1$, and $|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$. If $\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b $, then an angle between $\overrightarrow b $ and $\overrightarrow c $ is :
Let $\overrightarrow a = \widehat i + \widehat j - \widehat k$ and $\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$. Then the number of vectors $\overrightarrow b $ such that $\overrightarrow b \times \overrightarrow c = \overrightarrow a $ and $|\overrightarrow b | \in $ {1, 2, ........, 10} is :
If $\overrightarrow a \,.\,\overrightarrow b = 1,\,\overrightarrow b \,.\,\overrightarrow c = 2$ and $\overrightarrow c \,.\,\overrightarrow a = 3$, then the value of $\left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right),\,\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)} \right]$ is :
Let $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ ${a_i} > 0$, $i = 1,2,3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\overrightarrow a $ on the vector $3\widehat i + 4\widehat j$ be 7. Let $\overrightarrow b $ be a vector obtained by rotating $\overrightarrow a $ with 90$^\circ$. If $\overrightarrow a $, $\overrightarrow b $ and x-axis are coplanar, then projection of a vector $\overrightarrow b $ on $3\widehat i + 4\widehat j$ is equal to:
Let $\widehat a$ and $\widehat b$ be two unit vectors such that $|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2$. If $\theta$ $\in$ (0, $\pi$) is the angle between $\widehat a$ and $\widehat b$, then among the statements :
(S1) : $2|\widehat a \times \widehat b| = |\widehat a - \widehat b|$
(S2) : The projection of $\widehat a$ on ($\widehat a$ + $\widehat b$) is ${1 \over 2}$
Let $\widehat a$, $\widehat b$ be unit vectors. If $\overrightarrow c $ be a vector such that the angle between $\widehat a$ and $\overrightarrow c $ is ${\pi \over {12}}$, and $\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$, then ${\left| {6\overrightarrow c } \right|^2}$ is equal to :
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3$ and $|\vec{a} \times \vec{b}|^{2}=75$. Then $|\vec{a}|^{2}$ is equal to __________.
Explanation:
or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
$\therefore|\vec{b}|^{2}=6$
Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
$ 75=|\vec{a}|^{2} \cdot 6-9 $
$\therefore|\vec{a}|^{2}=14$
Let $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ be three non-coplanar vectors such that $\overrightarrow a $ $\times$ $\overrightarrow b $ = 4$\overrightarrow c $, $\overrightarrow b $ $\times$ $\overrightarrow c $ = 9$\overrightarrow a $ and $\overrightarrow c $ $\times$ $\overrightarrow a $ = $\alpha$$\overrightarrow b $, $\alpha$ > 0. If $\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right| = {1 \over {36}}$, then $\alpha$ is equal to __________.
Explanation:
Given,
$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $ ..... (i)
$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $ ..... (ii)
$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $ .... (iii)
Taking dot products with $\overrightarrow c ,\overrightarrow a ,\overrightarrow b $ we get
$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$
Hence,
(i) $ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$ ..... (iv)
(ii) $ \Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$ ..... (v)
(iii) $ \Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$ .... (vi)
Multiplying (iv), (v) and (vi)
$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha $ ..... (vii)
Dividing (vii) by (iv) $ \Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha $ ..... (viii)
Dividing (vii) by (v) $ \Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha $
Dividing (viii) by (vi) $ \Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$
Now, as given, $3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$
Let $\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = \widehat i + \widehat j + \widehat k$ and $\overrightarrow c $ be a vector such that $\overrightarrow a + \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow 0 $ and $\overrightarrow b \,.\,\overrightarrow c = 5$. Then the value of $3\left( {\overrightarrow c \,.\,\overrightarrow a } \right)$ is equal to _________.
Explanation:
Given: $\vec{a}+(\vec{b} \times \vec{c})=0$
$ \Rightarrow \vec{a} \cdot \vec{b}=0 $ ........(ii)
Equation (i) and equation (ii) are contradicting.
If $\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$, $\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$ and $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$ are coplanar vectors and $\overrightarrow a \,.\,\overrightarrow c = 5$, $\overrightarrow b \bot \overrightarrow c $, then $122({c_1} + {c_2} + {c_3})$ is equal to ___________.
Explanation:
$2{C_1} + {C_2} + 3{C_3} = 5$ ...... (i)
$3{C_1} + 3{C_2} + {C_3} = 0$ ...... (ii)
$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{ 2 & 1 & 3 \cr 3 & 3 & 1 \cr {{C_1}} & {{C_2}} & {{C_3}} \cr } } \right|$
$ = 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$
$ = 3{C_3} + 7{C_2} - 8{C_1}$
$ \Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$ ...... (iii)
${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$
So $122({C_1} + {C_2} + {C_3}) = 150$