The correct option is A a Let a=xi+yj+zk Now, #160; ( a.i )i+ ( a.j )j+ ( a.k )k=[(xi+yj+zk).i]i+[(xi+yj+zk).j]j+[(xi+yj+zk).k]k =xi+yj ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Applications of Dot Product

a.i i+ a.j j+...

Question

$(a . i) i + (a . j) j + (a . k) k =$

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $a$
Let $a = x i + y j + z k$
Now, $(a . i) i + (a . j) j + (a . k) k = [(x i + y j + z k) . i] i + [(x i + y j + z k) . j] j + [(x i + y j + z k) . k] k$
$= x i + y j + z k = a$

Suggest Corrections

Similar questions

i \times [(a - j) \times i] + j \times [(a - k) \times j] + k [(a - i) \times k] = 0

a = \frac{1}{2} (i + j + k)

(a . i) i + (a . j) j + (a . k) k = a

a

(a + b) x^{2} - 3 b x - 3 a = 0 \forall a, b \in R

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}} =

Solve for $x$ :

$3 (9^{x}) < 8 (3^{x}) + 3$

Solve for $x$ :

$3 (9^{x}) < 8 (3^{x}) + 3$

Join BYJU'S Learning Program