Question Video: Finding the Missing Value That Makes Two Vectors Parallel | Nagwa Question Video: Finding the Missing Value That Makes Two Vectors Parallel | Nagwa

# Question Video: Finding the Missing Value That Makes Two Vectors Parallel Mathematics • First Year of Secondary School

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If π = β©β, β +2βͺ and π =β©3β, 4β β 1βͺ, then one of the values of β that makes π β₯ π is οΌΏ. [A] 7 [B] 5 [C] β5 [D] β7

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### Video Transcript

If vector π is equal to β, β plus two and vector π is equal to three β, four β minus one, then one of the values of β that makes vector π parallel to vector π is what. Option (A) seven, option (B) five, option (C) negative five, or option (D) negative seven.

In this question, weβre given two vectors π and π. Their components include this unknown value of β. We need to find one of the values of β such that vector π and π will be parallel. So, letβs begin by recalling what it means to have two parallel vectors. In this case, as we have vector π and π, we would say that theyβre parallel if we can write that vector π is equal to π times vector π for some scalar quantity π, where π is not equal to zero. So, letβs set our two vectors as parallel vectors by saying that vector π is equal to π times vector π. We can then fill in the values of the vectors that weβre given. Vector π is β, β plus two, and vector π is three β, four β minus one.

On the right-hand side, we can multiply π by each of the π₯- and π¦-components. We can then evaluate the π₯-components. So we would have β is equal to three πβ. Dividing both sides by β, we have that one is equal to three π. Dividing three by three, we have one-third is equal to π or π is equal to one-third. Next, letβs consider the π¦-components. We can write that β plus two is equal to π times four β minus one. We have already established that π is equal to one-third, so letβs substitute this into the equation. We then have that β plus two is equal to one-third times four β minus one. Now letβs expand the parentheses on the right-hand side. One-third multiplied by four β will give us four-thirds β, and one-third multiplied by negative one gives us negative one-third.

We can then subtract β from both sides, remembering that if we have four-thirds β and we take away β, that will leave us with one-third β. We can then add one-third to both sides, remembering that two and one-third is equivalent to seven-thirds. Finally, we can multiply by three on both sides of this equation, which leaves us with seven is equal to one β or simply seven is equal to β. Therefore, one of the values of β that makes π parallel to π is seven. So, the answer is that given in option (A) seven. We can check this by plugging the value of seven into vectors π and π. When β is equal to seven, vector π will be seven, and seven plus two is nine. For vector π, when β is equal to seven, three times seven is 21 and four times seven is 28; subtract one will give us 27.

So, are vectors π and π scalar multiples? Yes, they are because we could write that vector π is equal to one-third of vector π. So, weβve confirmed that when β is seven, vectors π and π are parallel. But letβs check if any of the options given in (B), (C), or (D) for the value of β will also make the vectors π and π parallel. This time, weβll take the value given in the answer option for the value of β and then check if the vector π is parallel to vector π.

So, for option (B), weβre checking the value of β is equal to five. And vector π will therefore be five, and five plus two is seven. And vector π will be three times five, which is 15, and four times five is 20 minus one is 19. So, is there a scalar value π for which the vector five, seven is equal to π times the vector 15, 19? Well, no, thereβs not. If we look at the π₯-components, weβll have five is equal to 15π. That would mean that π is equal to one-third. However, π is equal to one-third would not fit the equation seven is equal to π times 19. So, that means when β is equal to five, these two vectors five, seven and 15, 19 are not parallel. So, we can eliminate option (B).

Letβs check option (C). This time weβre checking the value of β is equal to negative five. So, vector π will be negative five, negative three, and vector π will be negative 15, negative 21. Once again, there is no value of π for which vector π is equal to π times vector π. So, we can eliminate option (C) because when β is equal to negative five, these two vectors are not parallel. Finally then, in option (D), weβre checking the value of β is equal to negative seven. When we plug this value into the vectors π and π, once again, there is no value of π which makes negative seven, negative five and negative 21, negative 29 parallel. We have therefore eliminated options (B), (C), and (D), leaving us with the value of seven.

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