Question

show that vectors a vector is = two ICAP + 3 J cap + 6 k cap b vector is = 3 ICAP - 6 J cap + 2 k cap and C vector is =6 ICAP + 2 J cap - 3 k cap are mutually perpendicular​

Answer 1

Answer:

As given in question,

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

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

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

As we know that the dot product of two perpendicular vector is 0.

So, Dot product of $\vec{a}$ and $\vec{b}$

$(\vec{a}).(\vec{b})\ =\ (2\hat{i}\ +\ 3\hat{j}\ +\ 6\hat{k}).(3\hat{i}\ -\ 6\hat{j}\ +\2\hat{k})$

                            = 6 - 18 + 12

                            = 0

Hence, $\vec{a}$ and $\vec{b}$ are perpendicular to each other.

Dot product of $\vec{b}$ and $\vec{c}$

$(\vec{b}).(\vec{c})\ =\ (3\hat{i}\ -\ 6\hat{j}\ +\ 2\hat{k}).(6\hat{i}\ +\ 2\hat{j}\ -\3\hat{k})$

                            = 18 - 12 - 6

                            = 0

Hence, $\vec{b}$ and $\vec{c}$ are perpendicular to each other.

Dot product of $\vec{c}$ and $\vec{a}$

$(\vec{c}).(\vec{a})\ =\ (6\hat{i}\ +\ 2\hat{j}\ -\ 3\hat{k}).(2\hat{i}\ +\ 3\hat{j}\ +\ 6\hat{k})$

                            = 12 + 6 - 18

                            = 0

Hence, $\vec{c}$ and $\vec{a}$ are perpendicular to each other.

Hence, all three vectors are perpendicular to each other.

Answer 2

Explanation:

Answer:

As given in question,

\vec{a}\ =\ 2\hat{i}\ +\ 3\hat{j}\ +\ 6\hat{k}

a

= 2

i

^

+ 3

j

^

+ 6

k

^

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

c

= 6

i

^

+ 2

j

^

− 3

k

^

As we know that the dot product of two perpendicular vector is 0.

So, Dot product of \vec{a}

a

and \vec{b}

b

= 6 - 18 + 12

= 0

Hence, \vec{a}

a

and \vec{b}

b

are perpendicular to each other.

Dot product of \vec{b}

b

and \vec{c}

c

= 18 - 12 - 6

= 0

Hence, \vec{b}

b

and \vec{c}

c

are perpendicular to each other.

Dot product of \vec{c}

c

and \vec{a}

a

(\vec{c}).(\vec{a})\ =\ (6\hat{i}\ +\ 2\hat{j}\ -\ 3\hat{k}).(2\hat{i}\ +\ 3\hat{j}\ +\ 6\hat{k})(

c

).(

a

) = (6

i

^

+ 2

j

^

− 3

k

^

).(2

i

^

+ 3

j

^

+ 6

k

^

)

= 12 + 6 - 18

= 0

Hence, \vec{c}

c

and \vec{a}

a

are perpendicular to each other.

Hence, all three vectors are perpendicular to each other.