Let a-2 -3 -4 k- b-2 k- and C-6 -3 -k- then find d- which is perpendicular to both a- and b- and c-d-2

Let d=d_1î+d_2ĵ+d_3k̂ ∵ nbsp;d⊥ (a & b) ∴ nbsp;d.a=0 & d.b=0 ⇒ 2d_1 3d_2+4d_3=0 ...(1) d_1 d_2+2d_3=0 ...(2) Since, c.d=2 ⇒6d_1 3d_2+d_3= ...

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Multiplication with Vectors

Let a⃗=2 î-3 ...

Question

Let $\to a = 2^i - 3^j + 4^k, \to b =^i -^j + 2^k$ and $\to c = 6^i - 3^j +^k$ , then find $\to d$ which is perpendicular to both $\to a$ and $\to b$ and $\to c . \to d = 2$

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\to a = 2^i + 3^j - 4^k, \to b =^i +^j +^k

\to c = 4^i + 3^k

| \to a \times (\to b \times \to c) | =

\to a = 2^i + 3^j +^k

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

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

\to d

\to a \times \to b = \to d \times \to b, \to d \cdot \to c = 8

\to d . \to b

\to a = 2^i +^j - 3^k, \to b =^i - 2^j +^k

\to c = -^i +^j - 4^k

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

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

\to A = 2^i +^j

\to B = 3^j -^k

\to C = 6^i - 2^k

\to A - 2 \to B + 3 \to C

\to a =^i + 4^j + 2^k, \to b = 3^i - 2^j + 7^k

\to c = 2^i -^j + 4^k

\to d

\to a

\to b

\to c \cdot \to d = 15

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