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

Applications of Dot Product

Vectors a⃗,...

Question

Vectors $\to a, \to b$ and $\to c$ are of the same length and when taken pair-wise they form equal angles. If $\to a =^i +^j$ and $\to b =^j +^k$ , then find vector $\to c$ .

Open in App

Solution

According to question,
$\begin{matrix} ∣ ∣ ¯ ¯ ¯ a ∣ ∣ = ∣ ∣ ¯ ¯ b ∣ ∣ = ∣ ∣ ¯ ¯ c ∣ ∣ = \sqrt{2} A n g l e s : ¯ ¯ ¯ a^{\land} ¯ ¯ b = ¯ ¯ b^{\land} ¯ ¯ c = ¯ ¯ ¯ a^{\land} ¯ ¯ c \Rightarrow ¯ ¯¯ ¯ a . ¯ ¯ b = ¯ ¯ b . ¯ ¯ c = ¯ ¯ ¯ a . ¯ ¯ c [∴ ¯ ¯¯ ¯ a . ¯ ¯ b = ∣ ∣ ¯ ¯¯ ¯ a ∣ ∣ . ∣ ∣ ¯ ¯ b ∣ ∣ c o s θ \Rightarrow 1 = y + z = x + y [∴ ¯ ¯ b = x ˆ i + y ˆ j + z ˆ k \Rightarrow y + z = 1 - - - - (i) s i m i l e r l y, [∴ x^{2} + y^{2} + z^{2} = 2 x + y = 1 - - - - (i i) w e k n o w, x = z N o w, x^{2} + y^{2} + z^{2} = 2 x^{2} + (1 - x)^{2} + x^{2} = 2 3 x^{2} - 2 x - 1 = 0 3 x^{2} - 3 x + x - 1 = 0 3 x (x - 1) + 1 (x - 1) = 0 (3 x - 1) (x - 1) = 0 ∴ x = 1, x = \frac{- 1}{3} N o w f r o m, ¯ ¯ b = x ˆ i + y ˆ j + z ˆ k = \frac{- 1}{3} ˆ i + \frac{4}{3} ˆ j + \frac{- 1}{3} ˆ k a n d o t h e r v e c t o r : x = 1 ˆ i + ˆ k i s c o r r e c t A n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

\to a^{'} =^i +^j, \to b^{'} =^i +^j + 2^k

\to c^{'} = 2^i +^j -^k

\to a, \to b, \to c

\to b

\to c

(\to a, \to b, \to c

{\to a}^{'}, {\to b}^{'}, {\to c}^{'}

\to a =^i +^j +^k

\to b =^i -^j +^k

\to c =^i +^j -^k

\to a =^i -^j -^k

(\to a \times \to b) \times (\to c \times \to d)

If $\to a =^i -^j +^k a n d \to b =^j -^k,$ then find a vector $\to c$ such that $\to a \times \to c = \to b a n d \to a . \to c = 3.....$

\to a =^i +^j +^k, \to c =^j -^k

\to b

\to a \times \to b = \to c

\to a \cdot \to b = 3.

| \to b |

\to a =^i +^j -^k, \to b = -^i + 2^j + 2^k

\to c = -^i + 2^j -^k

\to a + \to b

\to b + \to c

Join BYJU'S Learning Program