Question

If $2(\vec a \times \vec c)+3(\vec b \times \vec c)=0$, where $\vec a=2\hat i-5\hat j+5\hat k$, $\vec b=\hat i-\hat j+3\hat k$ and $(\vec a-\vec b)\cdot\vec c=-97$, find $|\vec c \times \vec k|^2$.

Hint:

If $\vec p\times\vec q=0$, then the vectors are always parallel.

Answer 1

Correct Answer: 218

Step 1: Simplify the given vector equation

Given,

2(𝐚 × 𝐜) + 3(𝐛 × 𝐜) = 0

Using distributive property of cross product:

(2𝐚 + 3𝐛) × 𝐜 = 0

Hence, vector 𝐜 is parallel to (2𝐚 + 3𝐛).

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Step 2: Compute the vector (2𝐚 + 3𝐛)

𝐚 = (2, −5, 5),   𝐛 = (1, −1, 3)

2𝐚 + 3𝐛 = 2(2, −5, 5) + 3(1, −1, 3)

= (4, −10, 10) + (3, −3, 9)

= (7, −13, 19)

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Step 3: Express vector 𝐜

Since 𝐜 is parallel to (7, −13, 19),

𝐜 = λ(7, −13, 19)

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Step 4: Use the given dot product condition

Given,

(𝐚 − 𝐛) · 𝐜 = −97

𝐚 − 𝐛 = (2, −5, 5) − (1, −1, 3) = (1, −4, 2)

Substitute 𝐜:

(1, −4, 2) · λ(7, −13, 19) = −97

λ(7 + 52 + 38) = −97

λ(97) = −97

λ = −1

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Step 5: Find vector 𝐜

𝐜 = −1(7, −13, 19)

𝐜 = (−7, 13, −19)

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Step 6: Evaluate |𝐜 × 𝐤|²

𝐤 = (0, 0, 1)

𝐜 × 𝐤 =

| i   j   k |
| −7   13   −19 |
| 0   0   1 |

𝐜 × 𝐤 = (13, 7, 0)

|𝐜 × 𝐤|² = 13² + 7²

= 169 + 49

= 218

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Final Answer:

The required value is
218