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Given the vectors A = 2i + 3j − 2k and B = -3i - j + k, determine: 2A + 3B; |4A - B|

Mathematics

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4 parts
a(i)

−5i+3j−k-5\mathbf{i} + 3\mathbf{j} - \mathbf{k}

a(ii)

371\sqrt{371}

b
3
c

12113 units2\frac{1}{2}\sqrt{113} \text{ units}^2

Step-by-step solution

11 steps

Hapa kuna suluhisho la swali la 1:

Part a(i)· 3 steps

2A+3B2A + 3B;

Kutokana na vekta A=2i+3j−2kA = 2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} na B=−3i−j+kB = -3\mathbf{i} - \mathbf{j} + \mathbf{k}.

  1. Kokotoa $2A$.

    2A=2(2i+3j−2k)=4i+6j−4k2A = 2(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) = 4\mathbf{i} + 6\mathbf{j} - 4\mathbf{k}
  2. Kokotoa $3B$.

    3B=3(−3i−j+k)=−9i−3j+3k3B = 3(-3\mathbf{i} - \mathbf{j} + \mathbf{k}) = -9\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}
  3. Jumlisha matokeo.

    2A+3B=(4i+6j−4k)+(−9i−3j+3k)2A + 3B = (4\mathbf{i} + 6\mathbf{j} - 4\mathbf{k}) + (-9\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}) 2A+3B=(4−9)i+(6−3)j+(−4+3)k2A + 3B = (4-9)\mathbf{i} + (6-3)\mathbf{j} + (-4+3)\mathbf{k} 2A+3B=−5i+3j−k2A + 3B = \boxed{-5\mathbf{i} + 3\mathbf{j} - \mathbf{k}}

Part a(ii)

371\sqrt{371}

∣4A−B∣|4A - B|;

  1. Kokotoa $4A$.

    4A=4(2i+3j−2k)=8i+12j−8k4A = 4(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) = 8\mathbf{i} + 12\mathbf{j} - 8\mathbf{k}
  2. Kokotoa 4A−B4A - B.

    4A−B=(8i+12j−8k)−(−3i−j+k)4A - B = (8\mathbf{i} + 12\mathbf{j} - 8\mathbf{k}) - (-3\mathbf{i} - \mathbf{j} + \mathbf{k}) 4A−B=(8−(−3))i+(12−(−1))j+(−8−1)k4A - B = (8 - (-3))\mathbf{i} + (12 - (-1))\mathbf{j} + (-8 - 1)\mathbf{k} 4A−B=(8+3)i+(12+1)j+(−8−1)k4A - B = (8+3)\mathbf{i} + (12+1)\mathbf{j} + (-8-1)\mathbf{k} 4A−B=11i+13j−9k4A - B = 11\mathbf{i} + 13\mathbf{j} - 9\mathbf{k}
  3. Kokotoa ukubwa wa vekta.

    ∣4A−B∣=(11)2+(13)2+(−9)2|4A - B| = \sqrt{(11)^2 + (13)^2 + (-9)^2} ∣4A−B∣=121+169+81|4A - B| = \sqrt{121 + 169 + 81} ∣4A−B∣=371|4A - B| = \sqrt{371} ∣4A−B∣=371|4A - B| = \boxed{\sqrt{371}}

Part b3

Kutokana na vekta U=3i−4j+2kU = 3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k} ni perpendicular kwa vekta V=2i+Pj+3kV = 2\mathbf{i} + P\mathbf{j} + 3\mathbf{k}.

  1. Tumia sheria ya dot product kwa vekta perpendicular.

    Ikiwa vekta mbili ni perpendicular, dot product yao ni sifuri (U⋅V=0U \cdot V = 0).

    (3)(2)+(−4)(P)+(2)(3)=0(3)(2) + (-4)(P) + (2)(3) = 0
  2. Tatua kwa PP.

    6−4P+6=06 - 4P + 6 = 0 12−4P=012 - 4P = 0 4P=124P = 12 P=124P = \frac{12}{4} P=3P = \boxed{3}

Part c

12113 units2\frac{1}{2}\sqrt{113} \text{ units}^2

Tumia uchambuzi wa vekta kubainisha eneo la pembetatu ambayo pande zake mbili ni vekta P=2i+2j+kP = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} na Q=i+6j+2kQ = \mathbf{i} + 6\mathbf{j} + 2\mathbf{k}.

  1. Kokotoa cross product ya vekta PP na QQ.

    Eneo la pembetatu iliyoundwa na vekta mbili kama pande zake ni 12∣P×Q∣\frac{1}{2} |P \times Q|.

    P×Q=∣ijk221162∣P \times Q = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 1 & 6 & 2 \end{vmatrix} P×Q=i((2)(2)−(1)(6))−j((2)(2)−(1)(1))+k((2)(6)−(2)(1))P \times Q = \mathbf{i}((2)(2) - (1)(6)) - \mathbf{j}((2)(2) - (1)(1)) + \mathbf{k}((2)(6) - (2)(1)) P×Q=i(4−6)−j(4−1)+k(12−2)P \times Q = \mathbf{i}(4 - 6) - \mathbf{j}(4 - 1) + \mathbf{k}(12 - 2) P×Q=−2i−3j+10kP \times Q = -2\mathbf{i} - 3\mathbf{j} + 10\mathbf{k}
  2. Kokotoa ukubwa wa cross product.

    ∣P×Q∣=(−2)2+(−3)2+(10)2|P \times Q| = \sqrt{(-2)^2 + (-3)^2 + (10)^2} ∣P×Q∣=4+9+100|P \times Q| = \sqrt{4 + 9 + 100} ∣P×Q∣=113|P \times Q| = \sqrt{113}
  3. Kokotoa eneo la pembetatu.

    Eneo=12∣P×Q∣\text{Eneo} = \frac{1}{2} |P \times Q| Eneo=12113 units2\text{Eneo} = \boxed{\frac{1}{2}\sqrt{113} \text{ units}^2}
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