Algebra and Modulus of Vectors JEE Previous Year Question Bank

Q 1 :
Let the position vector of the vertices A, B and C of a triangle be $2 \hat{i} + 2 \hat{j} + \hat{k}$ , $\hat{i} + 2 \hat{j} + 2 \hat{k}$ and $2 \hat{i} + \hat{j} + 2 \hat{k}$ respectively. Let $l_{1}$ , $l_{2}$ and $l_{3}$ be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides AB, BC and CA respectively, then $l_{1}^{2} + l_{2}^{2} + l_{3}^{2}$ equals : [2024]

$\frac{1}{5}$
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{1}{4}$

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(3)

Position vector of $A = 2 \hat{i} + 2 \hat{j} + \hat{k}$

Position vector of $B = \hat{i} + 2 \hat{j} + 2 \hat{k}$

Position vector of $C = 2 \hat{i} + \hat{j} + 2 \hat{k}$

$\Rightarrow$ Coordinates of A, B and C of $△ A B C$ are A(2, 2, 1), B(1, 2, 2) and C(2, 1, 2)

$A B = \sqrt{{(1 - 2)}^{2} + {(2 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{2} units$

$B C = \sqrt{{(2 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 2)}^{2}} = \sqrt{2} units$

$A C = \sqrt{{(2 - 2)}^{2} + {(1 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{2} units$

$\Rightarrow$ $△ A B C$ is an equilateral triangle with side $\sqrt{2}$ units. In an equilateral triangle, orthocenter and centroid will be same.

Centroid, $G (x, y, z) \equiv G (\frac{2 + 1 + 2}{3}, \frac{2 + 2 + 1}{3}, \frac{2 + 2 + 1}{3}) \equiv G (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})$

Mid-point of AB is $D (\frac{3}{2}, 2, \frac{3}{2})$

$l_{1} = \sqrt{{(\frac{5}{3} - \frac{3}{2})}^{2} + {(\frac{5}{3} - 2)}^{2} + {(\frac{5}{3} - \frac{3}{2})}^{2}} = \sqrt{\frac{1}{36} + \frac{1}{9} + \frac{1}{36}} = \sqrt{\frac{1}{6}}$

$l_{1} = l_{2} = l_{3} = \sqrt{\frac{1}{6}}$

$∴ l_{1}^{2} + l_{2}^{2} + l_{3}^{2} = 3 \times \frac{1}{6} = \frac{1}{2}$

Q 2 :
The position vectors of the vertices A, B and C of a triangle are $2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ , $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $- \hat{i} + \hat{j} + 3 \hat{k}$ respectively. Let $l$ denotes the length of the angle bisector AD of $∠ B A C$ where D is on the line segment BC, then $2 l^{2}$ equals : [2024]

50
49
45
42

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(3)

In , $△ A B C$

Position vector of $A = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$

Position vector of $B = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$

Position vector of $C = - \hat{i} + \hat{j} + 3 \hat{k}$

Coordinates of the triangle ABC are A(2, –3, 3), B(2, 2, 3) and C(–1, 1, 3).

AD is the angle bisector of $∠ B A C$ .

$\Rightarrow B D : D C = A B : A C \Rightarrow D$ is the mid-point of the side BC.

Let the coordinates of D be (x, y, z).

By mid-point formula, $x = \frac{2 - 1}{2}, y = \frac{2 + 1}{2}, z = \frac{3 + 3}{2}$

$\Rightarrow x = \frac{1}{2}, y = \frac{3}{2} and z = 3$

So, $A D^{2} = {(2 - \frac{1}{2})}^{2} + {(- 3 - \frac{3}{2})}^{2} + {(3 - 3)}^{2} = \frac{90}{4}$

$\Rightarrow l^{2} = A D^{2} = \frac{90}{4}$ $∴ 2 l^{2} = 2 \times \frac{90}{4} = 45$

Q 3 :
Let O be the origin and the position vectors of A and B be $2 \hat{i} + 2 \hat{j} + \hat{k}$ and $2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ respectively. If the internal bisector of $∠ A O B$ meets the line AB at C, then the length of OC is [2024]

$\frac{2}{3} \sqrt{31}$
$\frac{3}{2} \sqrt{34}$
$\frac{2}{3} \sqrt{34}$
$\frac{3}{2} \sqrt{31}$

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(3)

$| \vec{O A} |$ $= \sqrt{4 + 4 + 1} = 3$

$| \vec{O B} |$ $= \sqrt{4 + 16 + 16} = 6$

$\frac{O A}{O B} = \frac{A C}{C B} = \frac{3}{6} = \frac{1}{2}$

$C \equiv (\frac{4 + 2}{3}, \frac{4 + 4}{3}, \frac{2 + 4}{3})$

$∴ C \equiv (2, \frac{8}{3}, 2)$

$| \vec{O C} |$ $= \sqrt{4 + \frac{64}{9} + 4} = \sqrt{\frac{136}{3}} = \frac{2 \sqrt{34}}{3}$ .

Q 4 :
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}, \vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$ and $\vec{a} + α \vec{b} + β \vec{c} = \vec{0}$ , then $α + β$ is equal to [2024]

35
–30
30
–25

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(1)

We have

$\vec{a} + 5 \vec{b} = λ \vec{c}, λ \in R$ ... (i)

$\vec{b} + 6 \vec{c} = μ \vec{a}, μ \in R \Rightarrow \vec{a} = \frac{\vec{b} + 6 \vec{c}}{μ}$

From (i), $\frac{\vec{b} + 6 \vec{c}}{μ} + 5 \vec{b} = λ \vec{c} \Rightarrow \vec{b} (\frac{1}{μ} + 5) = \vec{c} (λ - \frac{6}{μ})$

Since, $\vec{b}$ and $\vec{c}$ are non-collinear.

$∴ \frac{1}{μ} = - 5 and λ = \frac{6}{μ} \Rightarrow μ = \frac{- 1}{5} and λ = - 30$

$∴ From (i), \vec{a} + 5 \vec{b} - λ \vec{c} = 0$

$\Rightarrow \vec{a} + 5 \vec{b} + 30 \vec{c} = 0 \Rightarrow α = 5 and β = 30 ∴ α + β = 35$

Q 5 :
Let the position vectors of three vertices of a triangle be $4 \vec{p} + \vec{q} - 3 \vec{r}$ , $- 5 \vec{p} + \vec{q} + 2 \vec{r}$ and $2 \vec{p} - \vec{q} + 2 \vec{r}$ . If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4}$ and $α \vec{p} + β \vec{q} + γ \vec{r}$ respectively, then $α + 2 β + 5 γ$ is equal to: [2025]

1
3
6
4

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(2)

Given Orthocenter $= \frac{\vec{p} + \vec{q} + \vec{r}}{4}$

Circumcenter $= α \vec{p} + β \vec{q} + γ \vec{r}$

Centroid $= \frac{(4 \vec{p} + \vec{q} - 3 \vec{r}) + (- 5 \vec{p} + \vec{q} + 2 \vec{r}) + (2 \vec{p} - \vec{q} + 2 \vec{r})}{3}$

$= \frac{\vec{p} + \vec{q} + \vec{r}}{3}$

Since, centroid divides the line joining orthocenter and the circumcenter in the ratio of 2 : 1,

So, $2 \cdot (α \vec{p} + β \vec{q} + γ \vec{r}) + 1 \cdot (\frac{\vec{p} + \vec{q} + \vec{r}}{4}) = 3 \cdot (\frac{\vec{p} + \vec{q} + \vec{r}}{3})$

$\Rightarrow 8 (α \vec{p} + β \vec{q} + γ \vec{r}) = 3 (\vec{p} + \vec{q} + \vec{r})$

$\Rightarrow 8 α \vec{p} + 8 β \vec{q} + 8 γ \vec{r} = 3 \vec{p} + 3 \vec{q} + 3 \vec{r}$
On comparing, we get

$α = \frac{3}{8}, β = \frac{3}{8} and γ = \frac{3}{8}$

$∴ α + 2 β + 5 γ = \frac{3}{8} + 2 \times \frac{3}{8} + 5 \times \frac{3}{8} = \frac{3}{8} + \frac{6}{8} + \frac{15}{8} = \frac{24}{8} = 3$ .

Q 6 :
If the components of $\vec{a} = α \hat{i} + β \hat{j} + γ \hat{k}$ along and perpendicular to $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and $\frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$ , then $α^{2} + β^{2} + γ^{2}$ is equal to : [2025]

26
23
18
16

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4

(1)

Let ${\vec{a}}_{1}$ = Component of $\vec{a}$ along $\vec{b}$ and ${\vec{a}}_{2}$ = Component of $\vec{a}$ perpendicular to $\vec{b}$

$∴ {\vec{a}}_{1} = \frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and ${\vec{a}}_{2} = \frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$

$∵ \vec{a} = {\vec{a}}_{1} + {\vec{a}}_{2}$

$∴ \vec{a} = \frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k}) + \frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$

$= \frac{44}{11} \hat{i} + \frac{11}{11} \hat{j} - \frac{33}{11} \hat{k} = 4 \hat{i} + \hat{j} - 3 \hat{k}$

On comparing, we get $α = 4, β = 1, γ = - 3$

Hence, $α^{2} + β^{2} + γ^{2} = 16 + 1 + 9 = 26$ .

Q 7 :
Let the three sides of a triangle ABC be given by the vectors $2 \hat{i} - \hat{j} + \hat{k}$ , $\hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ . Let G be the centroid of the triangle ABC. Then $6 (| \vec{A G} |^{2} + | \vec{B G} |^{2} + | \vec{C G} |^{2})$ is equal to __________. [2025]

0%

(164)

We have in $△ A B C$ , $\vec{A B} + \vec{C A} = \vec{B C}$

Let PV of $\vec{A}$ be $\vec{0}$ then $\vec{A B} = \vec{B} - \vec{A}$

$\Rightarrow P . V . of \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$

$\vec{C A} = \vec{A} - \vec{C}$

P.V. of $\vec{C} = - \hat{i} + 3 \hat{j} + 5 \hat{k}$

Now, P.V. of $\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{1}{3} (\hat{i} + 2 \hat{j} + 6 \hat{k})$

Then $\vec{A G} = \frac{1}{3} (\hat{i} + 2 \hat{j} + 6 \hat{k})$

$\Rightarrow | \vec{A G} |^{2} = \frac{41}{9}$

$\Rightarrow \vec{B G} = (\frac{1}{3} - 2) \hat{i} + (\frac{2}{3} + 1) \hat{j} + (2 - 1) \hat{k}$

$\Rightarrow | \vec{B G} |^{2} = \frac{59}{9}$

$\Rightarrow \vec{C G} = (\frac{1}{3} + 1) \hat{i} + (\frac{2}{3} - 3) \hat{j} + (2 - 5) \hat{k}$

$\Rightarrow | \vec{C G} |^{2} = \frac{146}{9}$

$∴ 6 (| \vec{A G} |^{2} + | \vec{B G} |^{2} + | \vec{C G} |^{2}) = 6 (\frac{41}{9} + \frac{59}{9} + \frac{146}{9}) = 164$ .

Topic Question Set

Q 3 :
Let O be the origin and the position vectors of A and B be $2 \hat{i} + 2 \hat{j} + \hat{k}$ and $2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ respectively. If the internal bisector of $∠ A O B$ meets the line AB at C, then the length of OC is [2024]

0%

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Topic Question Set

Q 1 : Let the position vector of the vertices A, B and C of a triangle be 2i^+2j^+k^, i^+2j^+2k^ and 2i^+j^+2k^ respectively. Let l1, l2 and l3 be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides AB, BC and CA respectively, then l12+l22+l32 equals : [2024]

Q 2 : The position vectors of the vertices A, B and C of a triangle are 2i^–3j^+3k^, 2i^+2j^+3k^ and –i^+j^+3k^ respectively. Let l denotes the length of the angle bisector AD of ∠BAC where D is on the line segment BC, then 2l2 equals : [2024]

Q 3 : Let O be the origin and the position vectors of A and B be 2i^+2j^+k^ and 2i^+4j^+4k^ respectively. If the internal bisector of ∠AOB meets the line AB at C, then the length of OC is [2024]

Q 4 : Let a→, b→ and c→ be three non-zero vectors such that b→ and c→ are non-collinear. If a→+5b→ is collinear with c→, b→+6c→ is collinear with a→ and a→+αb→+βc→=0→, then α+β is equal to [2024]

Q 6 : If the components of a→=αi^+βj^+γk^ along and perpendicular to b→=3i^+j^–k^ respectively, are 1611(3i^+j^–k^) and 111(–4i^–5j^–17k^), then α2+β2+γ2 is equal to : [2025]

Q 7 : Let the three sides of a triangle ABC be given by the vectors 2i^–j^+k^, i^–3j^–5k^ and 3i^–4j^–4k^. Let G be the centroid of the triangle ABC. Then 6(|AG→|2+|BG→|2+|CG→|2) is equal to __________. [2025]

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Q 3 :
Let O be the origin and the position vectors of A and B be $2 \hat{i} + 2 \hat{j} + \hat{k}$ and $2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ respectively. If the internal bisector of $∠ A O B$ meets the line AB at C, then the length of OC is [2024]