Question

FIND A VECTOR WHOSE LENGTH IS 7 AND WHICH IS PERPENDICULAR TO EACH OF THE VECTORS A=2I-3J+6K AND B=I CAP+J CAP-K CAP

Answer 1

Answer: The vector is $\vec{C} = \dfrac{-21\hat i+56\hat j+35\hat k}{\sqrt{98}}$

Explanation:

Given that,

Length = 7

$\vec{A} = 2\hat i-3\hat j + 6\hat k$

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

We know that,

Let us consider required vector is $\vec{C}$.

Now, the vectors A and B are perpendicular to each

$\vec{A}\times\vec{B}=\begin{vmatrix}i &j &k \\2&-3 &6 \\ 1&1 &-1 \end{vmatrix}$

$\vec{A}\times\vec{B} = \hat i(3-6)-\hat j(-2-6)+\hat k(2+3)$

$\vec{A}\times\vec{B} = -3\hat i+8\hat j+5\hat k$

$|\vec{A}\times\vec{B}| = \sqrt{(-3)^{2}+  (8)^{2}+(5)^{2}}$

$|\vec{A}\times\vec{B}| = \sqrt{98}$

Now, the vector is

$\vec{C} = \dfrac{7(\vec{A}\times\vec{B})}{|\vec{A}\times\vec{B}|}$

$\vec{C} = \dfrac{-21\hat i+56\hat j+35\hat k}{\sqrt{98}}$

Hence, the required vector is $\vec{C} = \dfrac{-21\hat i+56\hat j+35\hat k}{\sqrt{98}}$.

Answer 2

"The magnitude of the vector is given as 7, and the vector perpendicular to the given three vector as A = 2 i + 3 j + 6 k and B = i + j – k. therefore the perpendicular vector will be C = A \times B.  

$\Rightarrow C = (2 i + 3 j + 6 k ) \times (i + j -k )$

$\Rightarrow C = - 3 i + 8 j + 5 k$.  

The magnitude of C will then be $C\quad =\quad \sqrt { 98 }$therefore the unit vector of 7 be considered along the vector C as U = vector C / magnitude of C  

$\Rightarrow U = - 3 i + 8 j + 5 k / \sqrt 98$

And also C = 7 U thereby,  

$7U\quad =\quad 7(\frac { 3i+8j+5k }{ \sqrt { 98 } } )\Rightarrow 7 U\quad$

$=\quad -21i+56j+\frac { 35k }{ \sqrt { 98 } }$

$\Rightarrow 7 7U\quad =\quad (\frac { -3 }{ \sqrt { 2 } } )i+(\frac { 8 }{ \sqrt { 2 } } )j+(\frac { 5 }{ \sqrt { 2 } } )k$

$\RightarrowTherefore the required vector = -3i+8j+\frac { 5k }{ \sqrt { 2 } }$"