Multiplication of vectors - class-XI

Using : $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (\vec{A} . \vec{C})  - \vec{C} (\vec{A}.\vec{B})$

If A dot B = 0 and A dot C = 0, then A is perpendicular to both B and C. Therefore, A must be parallel to the cross product B x C.

The unit vector perpendicular to A and B is (A x B) / |A x B|. A x B = (2i + j) x (i - j) = -2(i x j) + (j x i) = -2k - k = -3k. The unit vector is -k or k.

The scalar product of two vectors is defined as A dot B = |A||B| cos(theta). If the scalar product is zero and the magnitudes are non-zero, then cos(theta) must be zero, which occurs at 90 degrees.

($A$$\times $$B$) = - ($B$$\times $$A$) which are equal and opposite in direction.
Hence it will have angle in between $180^0$

$\vec{v}=6\hat{i}+2\hat{j}-2\hat{k}$

The cross product $\vec{A}\times\vec{B}$  is defined as a vector $\vec{C}$ that is perpendicular (orthogonal) to both A and B, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

The vector perpendicular to two vectors is found by their cross product. Calculating (2i + 2j - k) x (6i - 3j + 2k) gives (4-3)i - (4+6)j + (-6-12)k = i - 10j - 18k. The magnitude is sqrt(1^2 + (-10)^2 + (-18)^2) = sqrt(1 + 100 + 324) = sqrt(425) = 5*sqrt(17).

$(\bar A+\bar B)\times (\bar A-\bar B)$

Two vectors are parallel if one is a scalar multiple of the other. Here, B = 3A (12 = 34, 9 = 33, 3 = 3*1), so they are parallel.

Velocity v = dP/dt = 2Bt. Acceleration a = dv/dt = 2B. Since P = A + Bt^2, at any time, the velocity is in the direction of B. The acceleration is also in the direction of B. The angle between them is 0, not 45 degrees. This question appears to have a conceptual error or typo.

Given that,

  $ \vec{A}=2\hat{i}-\hat{j}+\hat{k} $

 $ \vec{B}=\hat{i}-2\hat{j}+\hat{k} $

Now, the projection  $\vec{A}$ on $\vec{B}$

  $ =\dfrac{\vec{A}\centerdot \vec{B}}{|\vec{B}|} $

 $ =\dfrac{5}{\sqrt{6}} $

Hence, this is the required solution

The resultant magnitude R = 1 for vectors of magnitude 2 and 3 implies the vectors are in opposite directions (3 - 2 = 1). If they are in opposite directions, the angle between them is 180 degrees. The magnitude of the cross product is |A||B| sin(theta). Since sin(180) = 0, the cross product is 0.

The component of A along B is (A dot B) / |B|. A = 2i + 3j, B = 5i - j. A dot B = (25) + (3-1) = 10 - 3 = 7. |B| = sqrt(5^2 + (-1)^2) = sqrt(26). Component = 7/sqrt(26).

The scalar triple product (A+B-C) dot ((A-B) x (B-C)) expands using properties of cross products. (A-B) x (B-C) = A x B - A x C - B x B + B x C = A x B - A x C + B x C. Dotting this with (A+B-C) results in terms like A dot (A x B) which are zero. The result is 0.

Since c and d are perpendicular, their dot product is zero: (a + 2b) . (5a - 4b) = 0. Expanding this using a^2 = 1, b^2 = 1, and a . b = cos(theta) gives 5 + 6(a . b) - 8 = 0, so a . b = 1/2, making theta = pi/3.

A vector is perpendicular to A if their dot product is zero. For A = 2i + 3j + 4k, check option D: (21) + (32) + (4*-2) = 2 + 6 - 8 = 0.

This is an identical duplicate of question 443269. Two vectors are perpendicular if their dot product equals zero. For A = 2i+3j+4k and option D (i+2j-2k), we calculate: 21 + 32 + 4*(-2) = 2 + 6 - 8 = 0. This confirms they are perpendicular.

$\vec {A} \times \vec {B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 1\ 4 & 3 & -2\end{vmatrix}$

$\hat {n}=\displaystyle \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\displaystyle \frac{\hat{i}+10\hat{j}+17\hat{k}}{\sqrt{390}}$

$\vec{x}=|\vec{B}|\hat{n}=\displaystyle \frac{\sqrt{29}(\hat{i}+10\hat{j}+17\hat{k})}{\sqrt{390}}$

$\vec A$ is perpendicular to $\vec B$ and $\vec A$ is perpendicular to $\vec C$. Thus $\vec A$ must lie along the direction of the cross product of $\vec B$ and $\vec C$.

Alternatively:
Given,
$\vec {A}.\vec {B}=0$
$\vec{A}.\vec{C}=0$
$ \Rightarrow \vec {A}.\vec {B} - \vec {A}.\vec {C}=\vec{A}( \vec{B} -\vec{C}) =0$
$ \Rightarrow \vec{A} \perp (\vec{B} -\vec{C})$
$\Rightarrow \vec{A} \parallel (\vec{B} \times \vec{C})$

The cross product of two vectors is zero if they are parallel or anti-parallel. A is along the x-axis (i). Vector B must be a multiple of i. -4i is parallel to i.

$\left( \vec { A } \times \vec { B }  \right) =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 5 & 7 & -3 \ 15 & 21 & a \end{matrix} \right| =0\quad or\quad 5a+45=0\quad or\quad or\quad 7a+63=0\quad \Rightarrow a=-9$

By definition, C = A x B is perpendicular to both A and B. It is not necessarily perpendicular to the sum or difference of A and B. However, C is definitely NOT perpendicular to itself (C x C = 0, but C dot C = |C|^2, which is non-zero). Option C is the incorrect statement.

(A + B) dot (A - B) = |A|^2 - |B|^2. A = 4i + 5j - 6k, |A|^2 = 16 + 25 + 36 = 77. B = 2i - 3j + 4k, |B|^2 = 4 + 9 + 16 = 29. 77 - 29 = 48.

For vectors to be parallel, their components must be proportional: 2/6 = 3/alpha = 4/beta. 1/3 = 3/alpha => alpha = 9. 1/3 = 4/beta => beta = 12.

$  \overrightarrow{A} \times \overrightarrow{B}=0   \overrightarrow{A}  \ and  \overrightarrow{B} $ are parallel.
$  \overrightarrow{B} \times \overrightarrow{C}=0  \Rightarrow  \overrightarrow{B}  \ and   \overrightarrow{C} $ are parallel 
 $\Rightarrow \vec{A} \parallel \vec{C}$

The vector perpendicular to the $i$ & $j$ plane would be along the unit vector $k$.
Another vector in $i$ & $j$ plane can be perpendicular to $\vec{F}$
$(3\widehat{i} +4\widehat{j}) \perp (4\widehat{i} -3\widehat{j}) $
But,  from the options only $7\widehat{k} \perp (4\widehat{i} -3\widehat{j})$

A vector $\displaystyle \vec{A}$ is parallel to an another vector $\displaystyle \vec{B}$ if it can be written as
$\displaystyle \vec{A}= m\vec{B}$ where $m$ is a constant.
Here, $\displaystyle \vec{A}=\left ( \hat{i}-\hat{j}+2\hat{k} \right )=\frac{1}{3}\left ( 3\hat{i}-3\hat{j}+6\hat{k} \right )$
or $\displaystyle \vec{A}=\frac{1}{3}\vec{B}$
This implies that $\vec A || \displaystyle \vec{B}$ and magnitude of $\displaystyle \vec{A}$ is $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

$\vec{a}\times\vec{b}=\begin{pmatrix}x _1\hat {i}+y _1\hat {j}\end{pmatrix}\times\begin{pmatrix}x _2\hat {i}+y _2\hat {j}\end{pmatrix}$

$\;\;\;\;\;\;\;\;\;\;\;=x _1y _2\hat {k}-x _2y _1\hat {k}=\vec{0}\Rightarrow x _1y _2=x _2y _1$

The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B} = \hat{n} |A| |B| \sin x $, where  $\hat{n}$ is the unit vector perpendicular to both A and B vectors and x is the angle between them.
Here in this question the vector product of $\vec{P1}$ and $\vec{P2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2) the $\sin$ of the angle between them is zero.
From the given options, the vector parallel to the given vector  $\hat{i}$ is  $-4\hat{i}$. 

$\displaystyle \displaystyle \overrightarrow A .  \overrightarrow B = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow B$
and, $\overrightarrow A .  \overrightarrow C = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow C$
Also, $\overrightarrow B \times \overrightarrow C $ is perpendicular to both $\vec { B } \ and  \ \vec { C }$
$Thus, \overrightarrow { A } ||, (\overrightarrow { B } \times \overrightarrow { C } )$

Magnitude of a vector A can be given as $ \sqrt{\overrightarrow{A}.\overrightarrow{A}} $. 
Hence, statement A is wrong .
Also, if $ \overrightarrow{a} = m \overrightarrow{b}$, taking dot product with b vector on both sides,
$ \overrightarrow{a}.\overrightarrow{b} = m   b^2 \Rightarrow m = \dfrac {\overrightarrow{a}.\overrightarrow{b}}{b^2} $
Hence , statement B is correct.

$\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$  and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$  is given. 

Since, the vector a is parallel to b, the corresponding coefficients of all the 3 components must bear the same ratio
i.e $ \dfrac{2}{1} = \dfrac{6n}{18} = \dfrac{m}{3} $
Or, $6n = 36, n = 6$
And $m = 6$

Vector C is perpendicular to both vectors A and B. Hence, it can be equal to their cross product.
Options A and B make vector C in the plane of vectors A  and B which is not possible.
In option D, this is not possible as vector B is not perpendicular to  vector A

Component of A along B = (A dot B / |B|^2) * B. A dot B = 2-1 = 1. |B|^2 = 1^2 + (-1)^2 = 2. Component = (1/2)(i - j). Perpendicular component = A - (component along B) = (2i + j) - (0.5i - 0.5j) = 1.5i + 1.5j = (3/2)(i + j).

Given,

In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

Simple example:- a=î,b=2î,c=–3î,d=0

(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c |  …. (#)

Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

This is the explanation of correct solution.