The problem involves vector operations, specifically cross product and dot product. Here are the steps to solve it:

Step 1: Given that p→×b→=c→×b→, we can say that p→ and c→ are in the same direction because the cross product of two vectors is another vector that is perpendicular to the plane containing the two original vectors. Therefore, we can write p→ = λc→, where λ is a scalar.

Step 2: We also know that p→⋅a→=0. Substituting p→ = λc→ into this equation, we get λc→⋅a→=0.

Step 3: Now, we can calculate c→⋅a→. c→ = iˆ−3jˆ+4kˆ and a→ = 3iˆ+jˆ−2kˆ. So, c→⋅a→ = (1*3) + (-3*1) + (4*-2) = 3 - 3 - 8 = -8.

Step 4: Substituting c→⋅a→ = -8 into λc→⋅a→=0, we get λ*(-8) = 0. Solving for λ, we find that λ = 0.

Step 5: Substituting λ = 0 into p→ = λc→, we find that p→ = 0.

Step 6: Finally, we need to find p→⋅(iˆ−jˆ−kˆ). Since p→ = 0, the dot product of p→ and any vector will be 0. Therefore, p→⋅(iˆ−jˆ−kˆ) = 0.

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The problem involves vector operations, specifically cross product and dot product. Here are the steps to solve it:

Step 1: Given that p→×b→=c→×b→, we can say that p→ and c→ are in the same direction because the cross product of two vectors is another vector that is perpendicular to the plane containing the two original vectors. Therefore, we can write p→ = λc→, where λ is a scalar.

Step 2: We also know that p→⋅a→=0. Substituting p→ = λc→ into this equation, we get λc→⋅a→=0.

Step 3: Now, we can calculate c→⋅a→. c→ = iˆ−3jˆ+4kˆ and a→ = 3iˆ+jˆ−2kˆ. So, c→⋅a→ = (1*3) + (-3*1) + (4*-2) = 3 - 3 - 8 = -8.

Step 4: Substituting c→⋅a→ = -8 into λc→⋅a→=0, we get λ*(-8) = 0. Solving for λ, we find that λ = 0.

Step 5: Substituting λ = 0 into p→ = λc→, we find that p→ = 0.

Step 6: Finally, we need to find p→⋅(iˆ−jˆ−kˆ). Since p→ = 0, the dot product of p→ and any vector will be 0. Therefore, p→⋅(iˆ−jˆ−kˆ) = 0.

Knowee AI · Accepted Answer

The problem involves vector operations, specifically cross product and dot product. Here are the steps to solve it:

Step 1: Given that p→×b→=c→×b→, we can say that p→ and c→ are in the same direction because the cross product of two vectors is another vector that is perpendicular to the plane containing the two original vectors. Therefore, we can write p→ = λc→, where λ is a scalar.

Step 2: We also know that p→⋅a→=0. Substituting p→ = λc→ into this equation, we get λc→⋅a→=0.

Step 3: Now, we can calculate c→⋅a→. c→ = iˆ−3jˆ+4kˆ and a→ = 3iˆ+jˆ−2kˆ. So, c→⋅a→ = (1*3) + (-3*1) + (4*-2) = 3 - 3 - 8 = -8.

Step 4: Substituting c→⋅a→ = -8 into λc→⋅a→=0, we get λ*(-8) = 0. Solving for λ, we find that λ = 0.

Step 5: Substituting λ = 0 into p→ = λc→, we find that p→ = 0.

Step 6: Finally, we need to find p→⋅(iˆ−jˆ−kˆ). Since p→ = 0, the dot product of p→ and any vector will be 0. Therefore, p→⋅(iˆ−jˆ−kˆ) = 0.

Let a⃗ =3iˆ+jˆ−2kˆ,b⃗ =4iˆ+jˆ+7kˆ and c→=iˆ−3jˆ+4kˆ be three vectors. If a vectors p→ satisfies p→×b→=c→×b→ and p→⋅a→=0, then p→⋅(iˆ−jˆ−kˆ) is equal to

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