Find a vector of magnitude $3$ and perpendicular to both the vectors $\to a = 2 ¯ i - 2 ¯ j + ¯ ¯ ¯ k$ and $¯ ¯ b = 2 ¯ i + 2 ¯ j + 3 ¯ ¯ ¯ k$

Open in App

Solution

Suppose a vector $\to c$ which has a magnitude $3$ and perpendicular to both the vectors $\to a$ and $\to b$
where $\to a = 2^i - 2^j +^k$ & $\to b = 2^i + 2^j + 3^k$
Now, a unit vector $\to d$ which is perpendicular to the vectors $\to a$ and $= \to b$ can be formed as
$\to d = \frac{\to a \times \to b}{| \to a \times \to b |}$
$\to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 2 & 1 2 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (- 6 - 2) -^j (6 - 2) +^k (4 + 4)$
$| \to a \times \to b | = \sqrt{(- 8)^{2} + (- 4)^{2} + (8)^{2}}$
$= \sqrt{64 + 16 + 64}$
$= \sqrt{144}$
$= 12$
$\to d = \frac{\to a \times \to b}{| \to a \times \to b} = \frac{8^i - 4^j + 8^k}{12} = \frac{- 2}{3}^i - \frac{1}{3}^j + \frac{2}{3}^k$
So, $\to c = 3 \to a$
$= 3 (\frac{- 2}{3}^i + \frac{1}{3}^j + \frac{2}{3}^k)$
$= - 2^i -^j + 2^k$
Hence, $\to c = - 2^i -^j + 2^k$

Suggest Corrections

Similar questions

3

b = 2 i - 2 j + k

c = 2 i + 2 j + 3 k

4 \hat{i} - \hat{j} + 3 \hat{k} and - 2 \hat{i} + \hat{j} - 2 \hat{k} .

\vec{a} = 2 \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .

2 i + 3 j - k

i - 2 j + 3 k

c \cdot (2 i - j + k) = - 6

\sqrt{2}

3 i - j - k

i + j - 2 k

2 i + 2 j + k

Join BYJU'S Learning Program