Let a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}, b⃗=2i^+2j^+k^\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k} and d⃗=a⃗×b⃗\vec{d} = \vec{a} \times \vec{b}. If c⃗\vec{c} is a vector such that a⃗⋅c⃗=∣c⃗∣\vec{a} \cdot \vec{c} = |\vec{c}|, ∣c⃗−2a⃗∣2=8|\vec{c} - 2 \vec{a}|^2 = 8 and the angle between d⃗\vec{d} and c⃗\vec{c} is π4\frac{\pi}{4}, then ∣10−3b⃗⋅c⃗∣+∣d⃗×c⃗∣2|10 - 3 \vec{b} \cdot \vec{c}| + |\vec{d} \times \vec{c}|^2 is equal to

Answer: 6

Step-by-step solution

Step 1: Calculate vectors a⃗\vec{a}, b⃗\vec{b}, and d⃗\vec{d} and their magnitudes

First, we calculate the magnitudes of vectors a⃗\vec{a} and b⃗\vec{b}. Then, we compute the cross product d⃗=a⃗×b⃗\vec{d} = \vec{a} \times \vec{b} using the determinant formula. Finally, we find the magnitude of vector d⃗\vec{d}. These values will be used in subsequent calculations.

Step 2: Determine the magnitude of vector c⃗\vec{c}

We use the given condition ∣c⃗−2a⃗∣2=8|\vec{c} - 2 \vec{a}|^2 = 8. Expanding the dot product, we get ∣c⃗∣2−4a⃗⋅c⃗+4∣a⃗∣2=8|\vec{c}|^2 - 4 \vec{a} \cdot \vec{c} + 4 |\vec{a}|^2 = 8. Substituting the given a⃗⋅c⃗=∣c⃗∣\vec{a} \cdot \vec{c} = |\vec{c}| and the calculated ∣a⃗∣=3|\vec{a}| = \sqrt{3}, we form a quadratic equation in terms of ∣c⃗∣|\vec{c}|. Solving this equation yields ∣c⃗∣=2|\vec{c}| = 2.

Step 3: Calculate b⃗⋅c⃗\vec{b} \cdot \vec{c}

We use the given conditions a⃗⋅c⃗=∣c⃗∣\vec{a} \cdot \vec{c} = |\vec{c}| and the angle between d⃗\vec{d} and c⃗\vec{c} is π4\frac{\pi}{4} to find d⃗⋅c⃗\vec{d} \cdot \vec{c}. We express c⃗\vec{c} as a linear combination of a⃗\vec{a}, d⃗\vec{d}, and a vector perpendicular to both, which is parallel to a⃗×d⃗\vec{a} \times \vec{d}. By substituting the known dot products and magnitudes, we solve for the scalar coefficient kk. This yields two possible values for kk, leading to two possible vectors for c⃗\vec{c} and consequently two possible values for b⃗⋅c⃗\vec{b} \cdot \vec{c}.

Step 4: Calculate ∣d⃗×c⃗∣2|\vec{d} \times \vec{c}|^2

We use the formula for the magnitude of the cross product: ∣d⃗×c⃗∣=∣d⃗∣∣c⃗∣sin⁡θ|\vec{d} \times \vec{c}| = |\vec{d}||\vec{c}|\sin\theta, where θ\theta is the angle between d⃗\vec{d} and c⃗\vec{c}. We substitute the calculated magnitudes of d⃗\vec{d} and c⃗\vec{c}, and the given angle π4\frac{\pi}{4}, to find ∣d⃗×c⃗∣2|\vec{d} \times \vec{c}|^2.

Step 5: Calculate the final expression for both cases

We substitute the two possible values for b⃗⋅c⃗\vec{b} \cdot \vec{c} and the calculated value of ∣d⃗×c⃗∣2|\vec{d} \times \vec{c}|^2 into the given expression. In both cases, the result is 6.

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