Let -3 - and -2 -3 k-If -1-2- where -1is parallel to - and -2 is perpendicular to - then -1-2 is equal to-B- frac 123 hat i -9 hat j -5 hat k C- -3 -9 -5 k-D- 3 -9 -5 k-

Α=3î+ĵ and β=2î ĵ+3k̂ So, β_1=λ(3î+ĵ) nbsp;and β_2= nbsp;β_1 β=λ(3î+ĵ) (2î ĵ+3k̂) =(3λ 2)î+(λ+1)ĵ 3k̂ Also, β_2·α=0 ⇒ (3λ 2)·3+(λ+1)·1+0=0 ⇒λ=1/2 ⇒β_ ...

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Byju's Answer

Standard XII

Mathematics

General Solution of cos theta = cos alpha

Let α⃗=3 î+ĵ ...

Question

$L e t \to α = 3^i +^j a n d \to β = 2^i -^j + 3^k . I f \to β =_{1} -_{2}, w h e r e_{1}$
is parallel to $\to α$ and ${\to β}_{2}$ is perpendicular to $\to α$ , then $_{1} \times_{2}$ is equal to:

\frac{1}{2} (- 3^i + 9^j + 5^k)

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\frac{1}{2} (3^i - 9^j + 5^k)

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- 3^i + 9^j + 5^k

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3^i - 9^j - 5^k

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Solution

The correct option is A $\frac{1}{2} (- 3^i + 9^j + 5^k)$
$\to α = 3^i +^j a n d \to β = 2^i -^j + 3^k S o,_{1} = λ (3^i +^j) a n d_{2} =_{1} - \to β = λ (3^i +^j) - (2^i -^j + 3^k) = (3 λ - 2)^i + (λ + 1)^j - 3^k A l s o,_{2} \cdot \to α = 0 \Rightarrow (3 λ - 2) \cdot 3 + (λ + 1) \cdot 1 + 0 = 0 \Rightarrow λ = \frac{1}{2} \Rightarrow_{1} = \frac{3}{2}^i + \frac{1}{2}^j \Rightarrow_{2} = - \frac{1}{2}^i + \frac{3}{2}^j - 3^k N o w,_{1} \times_{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k \frac{3}{2} & \frac{1}{2} & 0 - \frac{1}{2} & \frac{3}{2} & - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ =^i (- \frac{3}{2}) -^j (- \frac{9}{2}) +^k (\frac{9}{4} + \frac{1}{4}) = \frac{1}{2} (- 3^i + 9^j + 5^k)$

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Let →α=3^i+^j and →β=2^i−^j+3^k.If →β=→β1−→β2, where →β1 is parallel to →α and →β2 is perpendicular to →α, then →β1×→β2 is equal to:

$L e t \to α = 3^i +^j a n d \to β = 2^i -^j + 3^k . I f \to β =_{1} -_{2}, w h e r e_{1}$
is parallel to $\to α$ and ${\to β}_{2}$ is perpendicular to $\to α$ , then $_{1} \times_{2}$ is equal to: