Question Video: Evaluating Expressions Involving the Subtraction and Scalar Multiplication of Given Vectors in Three Dimensions | Nagwa Question Video: Evaluating Expressions Involving the Subtraction and Scalar Multiplication of Given Vectors in Three Dimensions | Nagwa

# Question Video: Evaluating Expressions Involving the Subtraction and Scalar Multiplication of Given Vectors in Three Dimensions Mathematics • Third Year of Secondary School

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If π = 9π’ β 7π£ + 8π€ and π = 8π’ β 3π£ + 4π€, find π β (2/3 π).

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### Video Transcript

If vector π is equal to nine π’ minus seven π£ plus eight π€ and vector π is equal to eight π’ minus three π£ plus four π€, find vector π minus two-thirds of vector π.

In this question, we will need to recall how to subtract vectors and also how to multiply a vector by a scalar. Letβs begin by calculating two-thirds of vector π.

In order to multiply any vector by a scalar, we simply multiply each of the components by that scalar. Two-thirds multiplied by eight is equal to sixteen-thirds. Therefore, two-thirds multiplied by eight π’ is sixteen-thirds π’. Multiplying two-thirds by negative three π£ gives us negative two π£. Finally, multiplying two-thirds by four π€ gives us eight-thirds π€. Multiplying vector π by the scalar two-thirds gives us sixteen-thirds π’ minus two π£ plus eight-thirds π€.

We need to subtract this vector from vector π. In order to subtract two vectors, we simply subtract the corresponding components. Nine π’ minus sixteen-thirds π’ is equal to eleven-thirds π’. This is because nine can be rewritten as twenty-seven thirds or 27 over three. Subtracting the π£-components gives us negative five π£, as negative seven π£ minus negative two π£ is the same as negative seven π£ plus two π£.

Finally, eight π€ minus eight-thirds π€ is equal to sixteen-thirds π€. We know this as we can rewrite eight as twenty-four thirds. And twenty-four thirds minus eight-thirds is equal to sixteen-thirds. Vector π minus two-thirds of vector π is therefore equal to eleven-thirds π’ minus five π£ plus sixteen-thirds π€.

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