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

Open in App

Solution

$\to b \times \to c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 2 & 1 2 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 8^i - 4^j + 8^k$
$∣ ∣ \to b \times \to c ∣ ∣ = \sqrt{{(- 8)}^{2} + {(- 4)}^{2} + {(8)}^{2}} = 12$
So, a vector of magnitude $3$ and perpendicular to both vector $b$ and $c$ is $\frac{\to b \times \to c}{∣ ∣ \to b \times \to c ∣ ∣}$
$= \frac{- 8^i - 4^j + 8^k}{12} \times 3 = - 2^i -^j + 2^k$

Suggest Corrections

Similar questions

3

\to a = 2 ¯ i - 2 ¯ j + ¯ ¯ ¯ k

¯ ¯ b = 2 ¯ i + 2 ¯ j + 3 ¯ ¯ ¯ k

2 i + 3 j - k

i - 2 j + 3 k

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

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} .

\sqrt{2}

3 i - j - k

i + j - 2 k

2 i + 2 j + k

Join BYJU'S Learning Program